Students at a liberal arts college study for an average of 10 hours per week with a standard deviation of 2 hours per week. The distribution of their study time happens to be uni-modal, symmetric and bell shaped. Approximately 68% of students study between 8 and B hours a week. What is the value of B? Select one:

Answer:

[tex]P(\bar X < 5000)=P(Z<4 (0.2533))=P(Z<1.0132)=0.845[/tex]

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variable that represent the mass of minerals of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(\mu,\sigma)[/tex]  

Where [tex]\mu=?[/tex] and [tex]\sigma=?[/tex]

From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

2) Solution to the problem

For this case we know this condition given :

[tex]P(X<5000)=0.6[/tex]

We can use the Z score given by this formula:

[tex]Z=\frac{X-\mu}{\sigma}[/tex]

And using this formula we got:

[tex]P(Z<\frac{5000-\mu}{\sigma})=0.6[/tex]

And we can find a value on the normal standard distribution that accumulates 0.6 of the are aon the left and 0.4 of the area on the right, on this case the value is Z=0.2533. And we can use the following excel code to find it :"=NORM.INV(0.6,0,1)"

So then we can do this:

[tex]0.2533=\frac{5000-\mu}{\sigma}[/tex]  (1)

By the other hand when we find the z score for the sample mean we have this:

[tex]Z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

And we want to find this probability:

[tex]P(\bar X < 5000)[/tex]

And if we use the z score formula we got:

[tex]P(Z< \frac{5000 -\mu}{\frac{\sigma}{\sqrt{16}}})=P(Z<\sqrt{16} \frac{5000-\mu}{\sigma})[/tex]  (2)

And replacing condition (1) into equation (2) we got:

[tex]P(Z<4 (0.2533))=P(Z<1.0132)=0.845[/tex]

And we can use the following excel code to find it: "=NORM.DIST(1.0132,0,1,TRUE)"

Answer:

Step-by-step explanation:

The test statistic, t, in this hypothesis testing scenario, is calculated by using the formula t = ([tex]\overline{X}[/tex] - μ₀) / (σ / √n). Substituting given values into the formula, the answer is b) -4.076.

In this hypothetical testing question, the test statistic, t, is calculated by dividing the difference between the sample mean and the null hypothesis mean by the standard error.

By dividing the standard deviation by the square root of the sample size, the standard error is determined. Here, the sample mean is 68.5, the null hypothesis mean is 74, the standard deviation is 6.4, and the sample size is 20.

The formula to calculate t is:

t = ([tex]\overline{X}[/tex] - μ₀) / (σ / √n)

Applying the given figures to the formula, the calculation is:

t = (68.5 - 74) / (6.4 / √20) = -4.076

So, the correct answer is b) -4.076.

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Answer:

(B) The citizens of the city, because they lose help they could have used to buy a home.

Step-by-step explanation:

Nul and alternative hypotheses are:

Type II error occurs when one fails to reject null hypothesis when the null hypothesis was wrong.

In this case Type II error happens when the conclusion is the rate of home ownership is not increasing after tax cut, where actually it is.

With this conclusion city council does not continue tax cut, and citizens of the city is harmed because they lose help they could have used to buy a home.

Final answer:

The citizens of the city would be harmed by a Type II error as they would miss out on the help to buy homes.

Explanation:

Who would be harmed by a Type II error?

(B) The citizens of the city, because they lose help they could have used to buy a home.

A Type II error in this context would harm the citizens of the city as they would miss out on the intended help in buying homes due to the failure to detect an increase in the rate of home ownership.

Answer:

The equation has zeroes at -4, 0 and 4 and is a minimum cubic degree or a 3 degree equation.

Step-by-step explanation:

The polynomial has roots at places where f(x) cuts the x axis.

The function cuts the x-axis at 3 points: 4, -4, and 0.

It cuts the axis at the point, x=0; x=4; x=-4

Therefore, the points equation has to be of the form, k*x*(x-4)*(x+4)*p(x)

where k is any arbitrary constant and p(x) is a polynomial of any degree depending on what the equation does in the region and (5,∞) and (-∞,-5).

Therefore, the equation has zeroes at -4, 0 and 4 and is a minimum cubic degree or a 3 degree equation.

The function f(x) = x // 2 (integer division) is an example of a function from N to N that is surjective because every natural number is covered, but not injective because different numbers can result in the same output.

To provide an example of a function f from the natural numbers N to N that is surjective but not injective, consider the function f(x) = x // 2, where '//' denotes integer division. For the function to be surjective, each element y in N must have at least one x such that f(x) = y. This is indeed the case here since for any y > 0, we can choose x = 2y or x = 2y + 1, and f(x) will equal y. To show that it is not injective, we can find two different numbers, x1 and x2, such that f(x1) = f(x2). For instance, if x1 = 4 and x2 = 5, both f(x1) and f(x2) equal 2, thus violating the definition of injectivity. Hence, f(x) = x // 2 is surjective because every y in N is an image of some x, but it is not injective because at least two different values in the domain map to the same value in the codomain.

Answer:

a) Sample mean: 48.71

Sample median: 48.3

Sample variance: 5.41

Sample standard deviation: 2.37

Step-by-step explanation:

a) Sample mean:

[tex]\bar{X}=\frac{1}{N} \sum X_i=\frac{1}{80}*3896.9=48.71[/tex]

Sample median: M=48.3

Note: I order the data increasingly and take the value N / 2 = 40. In this way there are 39 values above and 39 values below the median.

Sample variance:

[tex]s^2=\frac{1}{N-1}\sum (X_i-\bar X)^2 =(\frac{1}{80-1})*427.74=5.41[/tex]

Sample standard deviation

[tex]s=\sqrt{s^2}=\sqrt{5.41}=2.37[/tex]

b)

Answer: value of the speed, llv(t)ll = 4.0

Step-by-step explanation:

Given;

time interval t1 =5, t2= 7

Length of path ( distance) L = 8

Speed = distance travelled/ time taken

Speed = dL/dt

Speed = 8/(t2-t1) = 8/(7-5)

Speed = 8/2

Speed = 4.0

Since it's moving at constant speed the speed = 4.0

Answer:

[tex]n=\frac{0.3(1-0.3)}{(\frac{0.025}{1.96})^2}=1290.78[/tex]  

And rounded up we have that n=1291

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]

2) Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]

The margin of error for the proportion interval is given by this formula:  

[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]    (a)  

And on this case we have that [tex]ME =\pm 0.025[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex]   (b)  

And replacing into equation (b) the values from part a we got:

[tex]n=\frac{0.3(1-0.3)}{(\frac{0.025}{1.96})^2}=1290.78[/tex]  

And rounded up we have that n=1291

Answer:

B.75%

Step-by-step explanation:

Answer:

A. 50%

Step-by-step explanation:

about 50% b/c over half the people who play sports all get injuries

Answer:

First question:

Top 6%: 4.87 ounces

Bottom 6%: 4.75 ounces

Second question:

Top 7%: Score of 649.4.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

For the first problem, we have that:

[tex]\mu = 4.81, \sigma = 0.04[/tex]

Top 6%

The value of X when Z has a pvalue of 0.94. This is [tex]Z = 1.555[/tex]

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]1.555 = \frac{X - 4.81}{0.04}[/tex]

[tex]X - 4.81 = 1.555*0.04[/tex]

[tex]X = 4.8722[/tex]

Bottom 6%

The value of X when Z has a pvalue of 0.06. This is [tex]Z = 1.555[/tex]

For the second problem, we have that:

[tex]\mu = 496, \sigma = 109[/tex]

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-1.555 = \frac{X - 4.81}{0.04}[/tex]

[tex]X - 4.81 = -1.555*0.04[/tex]

[tex]X = 4.7477[/tex]

Top 7%

The value of X when Z has a pvalue of 0.93. This is [tex]Z = 1.475[/tex]

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]1.475 = \frac{X - 496}{104}[/tex]

[tex]X - 496 = 104*1.475[/tex]

[tex]X = 649.4[/tex]

Answer:

f(x) = 3[tex](\frac{1}{2})^{x}[/tex]

Step-by-step explanation:

hope it helps!

The function that represents a vertical stretch of an exponential function is f(x) = (3)^x.

The function that represents a vertical stretch of an exponential function is f(x) = (3)x.

In this function, the base of the exponential term is 3, and the exponent, x, determines the position on the graph. When the value of x increases, the function values also increase at an exponential rate.

For example, when x = 1, f(1) = (3)1 = 3. When x = 2, f(2) = (3)2 = 9. The function values double with each increase of x.

Answer:

a) [tex]t_{crit}=1.34[/tex]

b) [tex]t_{crit}=-1.33[/tex]

c) [tex]t_{crit}=\pm 2.11[/tex]

Step-by-step explanation:

Part a

[tex]\alpha=0.1[/tex] represent the significance level

df =15

Since is a right tailed test the critical value is given by:

[tex]t_{crit}=1.34[/tex]

And we can use the following excel code to find it: "=T.INV(0.9,15)"

Part b

[tex]\alpha=0.1[/tex] represent the significance level

n=20 represent the sample size

First we need to find the degrees of freedom given by:

[tex]df=n-1=20-1=19[/tex]

Since is a left tailed test the critical value is given by:

[tex]t_{crit}=-1.33[/tex]

And we can use the following excel code to find it: "=T.INV(0.1,19)"

Part c

[tex]\alpha=0.05[/tex] represent the significance level

n=18 represent the sample size

First we need to find the degrees of freedom given by:

[tex]df=n-1=18-1=17[/tex]

The value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]

Since is a two tailed tailed we have two critical values is given by:

[tex]t_{crit}=\pm 2.11[/tex]

And we can use the following excel code to find it: "=T.INV(0.025,17)"

Answer:

a) P(24) = 1025.

b) P(24) = 191.

Step-by-step explanation:

This population can be modeled by the following exponential model.

[tex]P(t) = P_{0}e^{rt}[/tex]

In which P(t) is the population after t hours, [tex]P_{0}[/tex] is the initial population and r is the decimal growth rate.

The initial number of bacteria in the culture is 28. This means that [tex]P_{0} = 28[/tex].

Population after 24 hours.

(a) No antibiotic is present, so the relative growth rate is 15%.

So r = 0.15.

[tex]P(t) = P_{0}e^{rt}[/tex]

[tex]P(24) = 28e^{0.15*24} = 1024.75[/tex]

(b) An antibiotic is present in the culture, so the relative growth rate is reduced to 8%.

So r = 0.08.

[tex]P(t) = P_{0}e^{rt}[/tex]

[tex]P(24) = 28e^{0.08*24} = 190.99[/tex]

Answer:

The probability that there are 8 occurrences in ten minutes is

option B. 0 .0771

Step-by-step explanation:

Given:

Random Variable = x

Mean number of occurrences in ten minutes is 5.3.

The probability of an occurrence is the same in any two time periods of an equal length

To Find:

The probability that there are 8 occurrences in ten minutes  = ?

Solution:

Let X be the number of  occurrences of the event X

[tex]X \sim {Pois} (\lambda)[/tex]

[tex]\lambda = E(X) = 5.3[/tex]

Possion of distribution is given by ,

[tex]P(X=x) = \frac{e^{- \lambda} \lambda^{x}}{x!}[/tex]

Substituting the values,

[tex]P(X=8) = \frac{e^{- 5.3} 5.3^{8}}{8!}[/tex]

[tex]P(X=8) = \frac{(0.004994) ( 622596.904)}{40320}[/tex]

[tex]P(X=8) = \frac{(3109.24894)}{40320}[/tex]

P(X=8) = 0.0771

To find the probability of 8 occurrences in ten minutes with a mean of 5.3, use the Poisson distribution formula to calculate the probability, resulting in approximately 0.0771.

Given: Mean number of occurrences in ten minutes = 5.3

Formula: Probability mass function of a Poisson distribution is given by: P(x) = (e^-λ * λ^x) / x!

Calculations: Plugging in the values, P(X=8) = (e^(-5.3) * 5.3^8) / 8! ≈ 0.0771

The probability that at least one route to work will still be open is [tex]\( \frac{209}{216} \)[/tex] when leaving.

To find the probability that at least one route to work will still be open when you leave for work, we need to find the probability that none of the streets will be closed within the next hour.

Let [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex] represent the events that Alabaster, Brillantine, and Clancy Streets are closed due to accidents within the next hour, respectively.

The probability of each street being closed within the next hour is:

[tex]\[ P(A) = \frac{25}{60} = \frac{5}{12} \][/tex]

[tex]\[ P(B) = \frac{50}{60} = \frac{5}{6} \][/tex]

[tex]\[ P(C) = \frac{40}{60} = \frac{2}{3} \][/tex]

Since the accidents on each street are independent events, the probability that all three streets remain open is the product of the probabilities that each street remains open:

[tex]\[ P(\text{})[/tex] = [tex]P(\neg A \cap \neg B \cap \neg C) = P(\neg A) \times P(\neg B) \times P(\neg C) ][/tex]

where [tex]\( \neg A \)[/tex], [tex]\( \neg B \)[/tex], and[tex]\( \neg C \)[/tex] represent the complementary events that the streets are not closed.

So, we have:

[tex]\[ P(\text{})[/tex] = [tex](1 - P(A)) \times (1 - P(B)) \times (1 - P(C)) ][/tex]

[tex]\[ = \left(1 - \frac{5}{12}\right) \times \left(1 - \frac{5}{6}\right) \times \left(1 - \frac{2}{3}\right) \][/tex]

[tex]\[ = \left(\frac{7}{12}\right) \times \left(\frac{1}{6}\right) \times \left(\frac{1}{3}\right) \][/tex]

[tex]\[ = \frac{7}{12} \times \frac{1}{6} \times \frac{1}{3} \]\[ = \frac{7}{216} \][/tex]

Therefore, the probability that at least one route to work will still be open when you leave for work is [tex]\( 1 - \frac{7}{216} = \frac{209}{216} \)[/tex].

Answer:

d. It can be biased by one or two extremely small or large values

Step-by-step explanation:

Specially in a small sample, a single measure can cause a large difference.

For example, you are selling yourself as a tutor, you have five students. 4 of them got good grades, but the other one got 0. The arithmetic mean is not going to be kind to your averages, in virtue of the outlier.

So the correct answer is:

d. It can be biased by one or two extremely small or large values

The arithmetic mean's disadvantage is that extreme values or outliers in the data set can significantly skew the mean. This makes the mean a less accurate reflection of the central tendency or 'average' of the data.

The correct answer to your question is "d. It can be biased by one or two extremely small or large values." The arithmetic mean, often simply called the 'mean', is a type of average most commonly used. It is calculated by adding all numbers in a set and then dividing by the quantity of numbers. While useful, it has a significant disadvantage in its sensitivity to extreme values, also referred to as 'outliers'. If there are one or two extremely large or small numbers in the data set, these can drastically affect the calculated mean, thus not accurately representing the central tendency of the overall data.

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Washington High School won the meet.

Johnson High School came in second with 159 points.

Difference between first and second was 3 points.

Step-by-step explanation:

At a particular swim meet, the details of the points which were awarded to the first three places finish are given.

To find the total number of points scored by each school is equal to sum of multiplying the respective points which were awarded to number of respective places finish.  

i.e. Washington High School had 10 first places finish, so total points for first place finish [tex]= 10 \times10=100[/tex] points.

Similarly, for second places finish = [tex]6\times 8=48[/tex]

and for third places finish = [tex]=2\times 7=14[/tex]

Therefore, Washington High School had total points = 100 + 48 + 14 = 162 points

By using this method, we can make the matrices (Please refer the below attachment).

From the matrices,

Total number of points:

Among these scores of four schools, 162 is the highest score. So, Washington High School won the meet.

Johnson High School came in second with 159 points.

Difference between first and second = 162 - 159 = 3 points.

Answer:

Step-by-step explanation:

The model [tex]N (t)[/tex], the number of planets found up to time [tex]t[/tex], as a Poisson process. So, the [tex]N (t)[/tex] has distribution of Poison distribution with parameter [tex](\lambda t)[/tex]

a)

The mean for a month is, [tex]\lambda = \frac{1}{3}[/tex] per month

[tex]E[N(t)]= \lambda t\\\\=\frac{1}{3}

(24)\\\\=8[/tex]

(Here. t = 24)

For Poisson process mean and variance are same,

[tex]Var[N (t)]= Var[N(24)]\\= E [N (24)]\\=8[/tex]

(Poisson distribution mean and variance equal)

The standard deviation of the number of planets is,

[tex]\sigma( 24 )] =\sqrt{Var[ N(24)]}=\sqrt{8}= 2.828

[/tex]

b)

For the Poisson process the intervals between events(finding a new planet) have  independent  exponential  distribution with parameter [tex]\lambda[/tex]. The  sum  of [tex]K[/tex] of these  independent exponential has distribution Gamma [tex](K, \lambda)[/tex].

From the given information, [tex]k = 6[/tex] and [tex]\lambda =\frac{1}{3}

[/tex]

Calculate the expected value.

[tex]E(x)=\frac{\alpha}{\beta}\\\\=\frac{K}{\lambda}

\\\\=\frac{6}{\frac{1}{3}}\\\\=18[/tex]

(Here, [tex]\alpha =k[/tex] and [tex]\beta=\lambda[/tex])                                                                      

C)

Calculate the probability that she will become eligible for the prize within one year.

Here, 1 year is equal to 12 months.

P(X ≤ 12) = (1/Г  (k)λ^k)(x)^(k-1).(e)^(-x/λ)

[tex]=\frac{1}{Г  (6)(\frac{1}{3})^6}(12)^{6-1}e^{-36}\\\\=0.2148696\\=0.2419\\=21.49%[/tex]

Hence, the required probability is 0.2149 or 21.49%

We can calculate the expected value and standard deviation of the number of planets Audrey will find in the next two years using probability and statistics. Using geometric distribution, we can also calculate the expected value and standard deviation of the amount of time until Audrey is eligible for a prize after finding her sixth new planet. Finally, we can calculate the probability of Audrey becoming eligible for the prize within one year using the binomial distribution formula.

To find the expected value and standard deviation of the number of planets Audrey will find in the next two years, we can use the concepts of probability and statistics. Since she finds one planet every three months on average, the expected value can be calculated by multiplying the average number of planets found per month (1/3) by the number of months in two years (24). The standard deviation can be calculated using the formula sqrt(n * p * q), where n is the number of trials, p is the probability of success, and q is the probability of failure.

To find the expected value and standard deviation of the amount of time until Audrey is eligible for the prize after finding her sixth new planet, we can use the concept of geometric distribution. The expected value can be calculated by taking the reciprocal of the probability of success (1/6) and the standard deviation can be calculated using the formula sqrt((1-p) / (p^2)).

To find the probability that Audrey will become eligible for the prize within one year, we can calculate the cumulative probability of finding six or more new planets in one year using the binomial distribution formula.

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Answer:

APR = 2.41% x 12 = 28.92%

Step-by-step explanation:

Her APR is 28.92%.

Sheila's APR is calculated by multiplying the monthly periodic rate of 2.41% by 12, yielding an APR of 28.92%.

The question refers to the process of calculating an Annual Percentage Rate (APR) from a given monthly periodic rate. Sheila's monthly periodic rate is 2.41%. The APR can be calculated by multiplying this monthly rate by the number of months in a year, which is 12.

To find Sheila's APR, we perform the following calculation:

APR = Monthly Periodic Rate × Number of Periods in a Year = 2.41% × 12 = 28.92%.

Hence, Sheila's APR is 28.92%.

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Answer:

Option A) any numerical value in an interval or collection of intervals

Step-by-step explanation:

Continuous Random Variable:

Option A) any numerical value in an interval or collection of intervals

A continuous random variable can take on any numerical value within a given range or collection of ranges, and it's a characteristic feature of it to take an infinite number of values in any interval. Some examples of this can be a person's height, time, temperature, and weight in physics.

A continuous random variable is a type of random variable that can assume any numerical value in a given interval or collection of intervals, making option a correct. This is in contrast to a discrete random variable, which can only take on a finite number of values. A key characteristic of continuous random variables is that they can take on an infinite number of values in any interval.

For example, the height of a person can be treated as a continuous variable, since it can theoretically take any value within a certain range, not just whole number values. The same applies to variables such as time, temperature, and weight in physics.

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Step-by-step explanation:

I took 3 indicated as 5 and its adjacent angle to be 3

<2 = <4

As <2 =<3( corresponding angle)

And <3 = <4 ( Vertically opp.angle)

hence <2 = <4

<2 =>110

so , <4 = 110

So, <4 =>110

As <4 and <5 form linear pair

So <4 + <5 =>180

<5 = 180 -110 =>70

As i took <5 as replacing angle to <3

So According to Question fig

<3 =>70

Hence proved

Answer:

6 mg of the metal needs to be added.

Step-by-step explanation:

Let the amount (in mg) of metal that needs to be added by y.

Therefore, the amount of silver in the above metal is 0.04y.

Prior to mixing, 11 mg of a metal contained 38% of silver (Given).

Therefore, the amount of silver before= [tex]\frac{38}{100}*11[/tex]= 4.18 mg

The total amount of silver after mixing, 4.18 + 0.04y mg

The total amount of metal after mixing, 11 + y mg

New percentage of silver = 26% .

Thus, [tex]\frac{4.18+0.04y}{11+y}*100= 26[/tex]

[tex](4.18 +0.04 y) * 100 = 26 *(11 +y)[/tex]

[tex]418+4y=286+26y[/tex]

[tex]132=22y[/tex]

y=6 mg

Therefore, the amount of metal that needs to be added is 6 mg.

Answer:

Null hypothesis:[tex]p\leq 0.5[/tex]  

Alternative hypothesis:[tex]p > 0.5[/tex]  

[tex]z=\frac{0.765 -0.5}{\sqrt{\frac{0.5(1-0.5)}{51}}}=3.785[/tex]  

[tex]p_v =P(z>3.785)=7.68x10^{-5}[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.01[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of boys is significantly higher than 0.5.  

Step-by-step explanation:

1) Data given and notation

n=51 represent the random sample taken

X=39 represent the number of boys

[tex]\hat p=\frac{39}{51}=0.765[/tex] estimated proportion of boys

[tex]p_o=0.5[/tex] is the value that we want to test

[tex]\alpha=0.01[/tex] represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that with this method, the probability of a baby being a boy is greater than 0.5.:  

Null hypothesis:[tex]p\leq 0.5[/tex]  

Alternative hypothesis:[tex]p > 0.5[/tex]  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.765 -0.5}{\sqrt{\frac{0.5(1-0.5)}{51}}}=3.785[/tex]  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided [tex]\alpha=0.01[/tex]. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

[tex]p_v =P(z>3.785)=7.68x10^{-5}[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.01[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of boys is significantly higher than 0.5.  

Answer:

The 90% confidence interval would be given by (57.006;62.994)  

Step-by-step explanation:

Previous concepts

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

[tex]\bar X=60[/tex] represent the sample mean  

[tex]\mu[/tex] population mean (variable of interest)  

[tex]\sigma=10[/tex] represent the population standard deviation  

n=30 represent the sample size  

Assuming the X follows a normal distribution  

[tex]X \sim N(\mu, \sigma=10)[/tex]

The sample mean [tex]\bar X[/tex] is distributed on this way:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]  

The confidence interval on this case is given by:

[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)

The next step would be find the value of [tex]\z_{\alpha/2}[/tex], [tex]\alpha=1-0.90=0.1[/tex],[tex]\alpha/2 =0.05[/tex] and [tex]z_\alpha/2=1.64[/tex]  

Using the normal standard table, excel or a calculator we see that:  

[tex]z_{\alpha/2}=1.64[/tex]

Since we have all the values we can replace:

[tex]60 - 1.64\frac{10}{\sqrt{30}}=57.006[/tex]  

[tex]60 + 1.64\frac{10}{\sqrt{30}}=62.994[/tex]  

So on this case the 90% confidence interval would be given by (57.006;62.994)  

Final answer:

To construct the 90% confidence interval for the average number of days it takes to find a job using the website's service, the formula for a confidence interval is applied using the given sample mean of 60 days, population standard deviation of 10 days, and a sample size of 30 customers. The calculated interval is between 57 and 63 days.

Explanation:

To construct a 90% confidence interval for the population mean number of days it takes to find a job using the website's service, we need to use the formula:

CI = \(\bar{x} \pm z*\frac{\sigma}{\sqrt{n}}\)

Where:

\(\bar{x}\) is the sample mean (60 days)

\(z\) is the z-score that corresponds to the desired confidence level (For 90%, \(z=1.645\) from the z-table)

\(\sigma\) is the population standard deviation (10 days)

\(n\) is the sample size (30 customers)

Plugging the values into the formula gives:

CI = 60 \pm 1.645 * (\frac{10}{\sqrt{30}})

Calculating the margin of error:

ME = 1.645 * (\frac{10}{\sqrt{30}}) \approx 3.00

Now compute the confidence interval:

CI = [60 - 3.00, 60 + 3.00]

CI = [57.00, 63.00]

So, we are 90% confident that the population mean number of days it takes to find a job using the website's service is between 57 and 63 days.

Mutliply his hourly rate by 1.5 to find his night pay:

30.40 x 1.5 = $45.60 per hour at night.

Multiply his rate by number of hours:

45.60 x 15 = $684

Answer:

Descriptive

Step-by-step explanation:

Statistics can be broadly classified into two main branches

i) Descriptive and ii) inferential

Descriptive statistics deal with values such as mean, standard deviation from the data.

Inferential statistics is used to predict unknown values from the observed values.

Applied statistics mainly analyses the data according to the needs of business or industries.

Hence the average expenditure on different items such as food, clothes, fuel comes under

Descriptive Statistics

Final answer:

Average expenditure on different items like food, clothes, and fuel falls under descriptive statistics, essential for summarizing trends in expenditure data. The correct option is a.

Explanation:

Descriptive statistics involve summarizing trends in data, such as calculating averages and standard deviations, making them suitable for analyzing average expenditures on items like food and clothes.

Inferential statistics are used to make inferences about populations based on sample data, aiming to find cause and effect relationships or correlations, which is essential when studying average expenditures across different categories.

Therefore, analyzing average expenditures on various items falls under the domain of descriptive statistics as it involves summarizing and interpreting trends in expenditure data.

If a generic interval [tex][a,b][/tex] is partitioned into [tex]n[/tex] subintervals, each one has a length:

[tex]\Delta x = \dfrac{b-a}{n}.[/tex]

In this case, [tex]a = 4[/tex], [tex]b=6[/tex] and [tex]n=4[/tex], so:

[tex]\Delta x = \dfrac{6-4}{4} = \dfrac{2}{4} = \dfrac{1}{2} = 0.5.[/tex]

The grid points are given by:

[tex]x_k = a + k\Delta x, \quad\textrm{with } k \in \{0,1,2,3,4\}.[/tex]

Since [tex]a = 4[/tex] and [tex]\Delta x = 0.5[/tex], we have:

[tex]x_0 = 4 + 0 \times 0.5 = 4 \\x_1 = 4 + 1 \times 0.5 = 4 + 0.5 = 4.5\\x_2 = 4 + 2 \times 0.5 = 4 + 1 = 5\\x_3 = 4 + 3 \times 0.5 = 4 + 1.5 = 5.5\\x_4 = 4 + 4 \times 0.5 = 4 + 2 = 6[/tex]

The [tex]4[/tex] subintervals are of the form [tex]I_k = [x_{k-1}, x_k], \quad\textrm{with } k \in \{1, 2, 3,4\}[/tex]:

[tex]I_1 = [x_0, x_1] = [4,4.5]\\I_2 = [x_1, x_2] = [4.5,5]\\I_3 = [x_2, x_3] = [5,5.5]\\I_4 = [x_3, x_4] = [5.5, 6][/tex]

For the left Riemann sums we will use the left-handed points, namely:

[tex]\{x_0, x_1, x_2, x_3\} = \{4,4.5,5,5.5\}.[/tex]

For the right Riemann sums we will use the right-handed points, namely:

[tex]\{x_1, x_2, x_3, x_4\} = \{4.5,5,5.5,6\}.[/tex]

For the midpoint Riemann sums we will use the average between the two extrema of each subinterval, given by

[tex]I_k \to \tilde{x}_k = \dfrac{x_{k-1}-x_k}{2}, \quad \textrm{with } k \in\{1,2,3,4\}.[/tex]

This gives the midpoints:

[tex]I_1 = [x_0, x_1] = [4,4.5] \to \tilde{x}_1 = \dfrac{x_0 + x_1}{2} = \dfrac{4+4.5}{2} = \dfrac{8.5}{2} = 4.25\\\\I_2 = [x_1, x_2] = [4.5,5] \to \tilde{x}_2 = \dfrac{x_1 + x_2}{2} = \dfrac{4.5+5}{2} = \dfrac{9.5}{2} = 4.75\\\\I_3 = [x_2, x_3] = [5,5.5] \to \tilde{x}_3 = \dfrac{x_2 + x_3}{2} = \dfrac{5+5.5}{2} = \dfrac{10.5}{2} = 5.25\\\\I_4 = [x_3, x_4] = [5.5,6] \to \tilde{x}_4 = \dfrac{x_3 + x_4}{2} = \dfrac{5.5+6}{2} = \dfrac{11.5}{2} = 5.75[/tex]

The points used for the midpoint Riemann sums are therefore:

[tex]\{\tilde{x}_1,\tilde{x}_2,\tilde{x}_3,\tilde{x}_4\} =\{4.25,4.75,5.25,5.75\}.[/tex]

The interval and sub intervals are illustrations of Riemann sums

The given parameters are:

[tex]\mathbf{Interval = [4,6]}[/tex]

[tex]\mathbf{n =4}[/tex] --- sub intervals

(a) The sub interval length

This is calculated as:

[tex]\mathbf{\Delta x = \frac{b - a}{n}}[/tex]

Where:

[tex]\mathbf{[a,b] =[4,6]}[/tex]

So, we have:

[tex]\mathbf{\Delta x = \frac{6 - 4}{4}}[/tex]

[tex]\mathbf{\Delta x = \frac{2}{4}}[/tex]

[tex]\mathbf{\Delta x = 0.5}[/tex]

Hence, the length of the sub-intervals is 0.5

(b) The grid points

This is calculated as:

[tex]\mathbf{x_k = a + k\Delta x}[/tex]

So, we have:

[tex]\mathbf{x_0 = 4 + 0 \times 0.5 = 4}[/tex]

[tex]\mathbf{x_1 = 4 + 1 \times 0.5 = 4.5}[/tex]

[tex]\mathbf{x_2 = 4 + 2 \times 0.5 = 5}[/tex]

[tex]\mathbf{x_3 = 4 + 3 \times 0.5 = 5.5}[/tex]

[tex]\mathbf{x_4 = 4 + 4 \times 0.5 = 6}[/tex]

So, the grid points are 4, 4.5, 5, 5.5 and 6

(c) The left, right and midpoint Riemann sums

The midpoint is the average of the left and right points.

So, we have:

[tex]\mathbf{x_0 = 0.5 \times (4 + 4.5) = 4.25}[/tex]

[tex]\mathbf{x_1 = 0.5 \times (4.5 + 5) = 4.75}[/tex]

[tex]\mathbf{x_2 = 0.5 \times (5 + 5.5) = 5.25}[/tex]

[tex]\mathbf{x_3 = 0.5 \times (5.5 + 6) = 5.75}[/tex]

So, the midpoints are 4.25, 4.75, 5.25 and 5.75

Read more about Riemann sums at:

https://brainly.com/question/23960718

Answer: 1.7 x 10^6

Step-by-step explanation:

6^8 = 1,679,616

1,679,616 = 1.7 x 10^6

Answer: B. 264

Step-by-step explanation:

Formula to calculate the sample size 'n' , if the prior estimate of the population proportion (p) is available:

[tex]n= p(1-p)(\dfrac{z}{E})^2[/tex]

, where z = Critical z-value corresponds to the given confidence interval

E=  margin of error

Let p be the population proportion of clear days.

As per given , we have

Prior sample size : n= 150

Number of clear days in that sample = 117

Prior estimate of the population proportion of clear days = [tex]p=\dfrac{117}{150}[/tex]

E= 0.05

The critical z-value corresponding to 95% confidence interval = z*= 1.95 (By z-table)

Then, the required sample size will be :

[tex]n= \dfrac{117}{150}(1-\dfrac{117}{150})(\dfrac{1.96}{0.05})^2[/tex]

Simplify ,

[tex]n= (0.1716)(39.2)^2[/tex]

[tex]n= 263.687424\approx264[/tex]

Hence, the sample size necessary to construct this interval =264

Thus the correct option is B. 264