Answer:
[tex]P(x \geq 9)=P(X=9)+P(X=10)[/tex]
[tex]P(X=9)=(10C9)(0.9)^9 (1-0.9)^{10-9}=0.387[/tex]
[tex]P(X=10)=(10C10)(0.9)^{10} (1-0.9)^{10-10}=0.349[/tex]
And adding we got:
[tex]P(x \geq 9)=P(X=9)+P(X=10)=0.387+0.349=0.736[/tex]
Step-by-step explanation:
Previous concepts
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Solution to the problem
Let X the random variable of interest, on this case we now that:
[tex]X \sim Binom(n=10, p=1-0.1=0.9)[/tex]
Where 1-p = 1-0.1=0.9 represent the probability of being NOT defective
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
And we want to find this probability:
[tex]P(x \geq 9)=P(X=9)+P(X=10)[/tex]
[tex]P(X=9)=(10C9)(0.9)^9 (1-0.9)^{10-9}=0.387[/tex]
[tex]P(X=10)=(10C10)(0.9)^{10} (1-0.9)^{10-10}=0.349[/tex]
And adding we got:
[tex]P(x \geq 9)=P(X=9)+P(X=10)=0.387+0.349=0.736[/tex]
Add and simplify. Please help
Answer: Third answer
Step-by-step explanation:
11. Simplify the complex fraction. n-4/n² - 2n-15/n +1/n +3
NEED HELP ASAP!!!!
In simplifying complex fractions, first, put the individual fractions in simplest terms. Solve the equations based on the order of operations, simplify the fractions, find a common denominator, and combine the fractions. Your skill in Algebraic operations, particularly with polynomials and fractions, play a significant role in solving the problem.
Explanation:When it comes to simplifying complex fractions, you need to first ensure that the individual fractions are in their simplest terms. For the fraction n-4/n², consider factoring out n from the denominator, while for 2n-15/n +1/n +3, the expression can be simplified to (2n-15+n+3)/n or (3n-12)/n.
Following an order of operations, evaluate the expressions and perform the indicated operations for a simple and correct answer. Briefly, your steps should include: simplifying the complex fractions, finding a common denominator, and combining the fractions. Once you have done this, continue to simplify until you are left with one fraction.
Also note that a key aspect to solving this is applying your knowledge of Algebraic Operations, specifically with polynomials and fractions. It's important to take care not to make any mistakes in signs or operations as they can significantly alter your responses.
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The simplified form is (3n² - n + 11)/n².
To simplify the complex fraction (n - 4)/(n²) - (2n - 15)/n + 1/n + 3, follow these steps:
Find a common denominator for all the fractions involved. Here, the common denominator will be n².
1. Rewrite each term with the common denominator:
(n - 4)/n²
(2n - 15)/n becomes (2n - 15)n/n²
1/n becomes n/n²
3 becomes 3n²/n²
2. Combine all terms to have the same denominator:
(n - 4 - 2n + 15 + n + 3n²)/n²
3. Simplify the numerator:
(n - 4 - 2n + 15 + n + 3n²)
Combine like terms: (3n² + n - 2n + n - 4 + 15)
4. Simplifies to: (3n² - n + 11)
5. Rewrite the fraction:
(3n² - n + 11)/n²
So, the simplified fraction is (3n² - n + 11)/n².
Given the following winning percentages of the teams in a league (for a single year) compute the within-season standard deviation for the league. Team Season Winning Percentage 1 0.750 2 0.750 3 0.200 4 0.600 5 0.200 (a) 0.79 (b) 0.251 (x) 0.56 (d) 0.063
Answer:
(b) 0.251
Step-by-step explanation:
1. Standard deviation equation:
[tex]SD=\sqrt{\frac{\sum\limits^N_i {(x_{i}-X)^{2} } }{N} }[/tex]
Where X is the mean of the data, and N the amount of data. Then, N=5
2. Estimate the Mean:
[tex]X=\frac{0.750+0.750+0.200+0.600+0.200}{5}=\frac{2.5}{5}=0.5\\[/tex]
3. Caclulate Standard deviation:
[tex]SD=\sqrt{\frac{{(0.750-0.500)^{2}+(0.750-0.500)^{2}+(0.200-0.500)^{2}+(0.600-0.500)^{2}+(0.200-0.500)^{2} } }{5} }[/tex]
[tex]SD=\sqrt{\frac{0.315}{5} }=\sqrt{0.063}\\SD=0.251[/tex]
Suppose that X has a Poisson distribution with a mean of 64. Approximate the following probabilities. Round the answers to 4 decimal places (e.g. 98.7654). a) Upper P left-parenthesis Upper X greater-than 84right-parenthesis equals b) Upper P left-parenthesis Upper X less-than 64 right-parenthesis equals
Answer:
(a) The probability of the event (X > 84) is 0.007.
(b) The probability of the event (X < 64) is 0.483.
Step-by-step explanation:
The random variable X follows a Poisson distribution with parameter λ = 64.
The probability mass function of a Poisson distribution is:
[tex]P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0, 1, 2, ...[/tex]
(a)
Compute the probability of the event (X > 84) as follows:
P (X > 84) = 1 - P (X ≤ 84)
[tex]=1-\sum _{x=0}^{x=84}\frac{e^{-64}(64)^{x}}{x!}\\=1-[e^{-64}\sum _{x=0}^{x=84}\frac{(64)^{x}}{x!}]\\=1-[e^{-64}[\frac{(64)^{0}}{0!}+\frac{(64)^{1}}{1!}+\frac{(64)^{2}}{2!}+...+\frac{(64)^{84}}{84!}]]\\=1-0.99308\\=0.00692\\\approx0.007[/tex]
Thus, the probability of the event (X > 84) is 0.007.
(b)
Compute the probability of the event (X < 64) as follows:
P (X < 64) = P (X = 0) + P (X = 1) + P (X = 2) + ... + P (X = 63)
[tex]=\sum _{x=0}^{x=63}\frac{e^{-64}(64)^{x}}{x!}\\=e^{-64}\sum _{x=0}^{x=63}\frac{(64)^{x}}{x!}\\=e^{-64}[\frac{(64)^{0}}{0!}+\frac{(64)^{1}}{1!}+\frac{(64)^{2}}{2!}+...+\frac{(64)^{63}}{63!}]\\=0.48338\\\approx0.483[/tex]
Thus, the probability of the event (X < 64) is 0.483.
To approximate probabilities for a Poisson distribution with mean 64, one can use the cumulative distribution function of the Poisson distribution. But for large means like 64, using a Normal approximation to the Poisson might be more accurate.
Explanation:The question is about the Poisson distribution which is a probability distribution used often in statistical modelling where we are counting the number of times an event happens in a fixed interval of time or space. The mean is given as 64. To compute probabilities for values more than or less than a certain value, we use the cumulative distribution function (cdf) of the Poisson distribution.
The cumulative distribution function gives us the probability that the random variable is less than or equal to a certain value. In other words, if we want to find P(X > 84), we can find it by subtracting P(X <= 83) from 1. Likewise, to find P(X < 64), we can directly use the cdf at 63.
Please note however, because of the large mean (64), it may be more accurate to use a Normal approximation to the Poisson with mean 64 and variance 64. In that case, standardize the variables and use the Standard Normal table for the probabilities.
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In a survey of 5100 randomly selected T.V. viewers, 40% said they watch network news programs. Find the margin of error for this survey if we want 95% confidence in our estimate of the percent of T.V. viewers who watch network news programs.
Answer:
Margin of error = 0.01344
Step-by-step explanation:
Margin of error = critical value × standard deviation.
critical value for 95% confidence interval = 1.96
Standard deviation = √[(p)(q)/n]
p = 0.4, q = 1 - p = 1 - 0.4 = 0.6, n = 5100
Standard deviation = √[(0.6×0.4)/5100] = 0.00686
Margin of error = 1.96 × 0.00686 = 0.0134
The probability a student at a university is a business major is 0.17. The probability a student at a university is receiving financial aid is 0.55. Assume whether is a student is a business major is independent of whether the student is receiving financial aid. What is the probability the student is a business major and receiving financial aid
Answer: 0.0935
Step-by-step explanation: p(business major) = 0.17, p(receiving financial aids) = 0.55
The two events are independent that's the occurrence of one does not affect the other.
Hence, the probability of that the student is a business major and receiving financial aid is given below as
p(business major and financial aid) = 0.17×0.55 = 0.0935.
An art history professor assigns letter grades on a test according to the following scheme. A: Top 5% of scores B: Scores below the top 5% and above the bottom 62% C: Scores below the top 38% and above the bottom 22% D: Scores below the top 78% and above the bottom 5% F: Bottom 5% of scores Scores on the test are normally distributed with a mean of 73.3 and a standard deviation of 9.7. Find the numerical limits for a B grade. Round your answers to the nearest whole number, if necessary.
Answer:
76 ≤ B ≤ 89
Step-by-step explanation:
Mean score (μ) = 73.3
Standard deviation (σ) = 9.7
If B Scores are below the top 5% and above the bottom 62%, then:
62% ≤ B ≤ 95%
In a normal distribution, the 62nd percentile has a corresponding z-score of z = 0.305, while the 95th percentile has a corresponding z-score of z = 1.650.
The grades X1 and X2 which are the limits for a B grade are given by:
[tex]z= \frac{X-\mu}{\sigma}\\0.305= \frac{X_1-73.3}{9,7}\\X_1=76.2\\1.650= \frac{X_2-73.3}{9,7}\\X_2=89.3[/tex]
Rounding to the nearest whole number, a B grade is given to grades between 76 and 89.
Final answer:
To calculate the numerical limits for a B grade in a normally distributed set of test scores with a mean of 73.3 and a standard deviation of 9.7, we find the z-scores corresponding to the top 5% and bottom 62%. The B grade range is approximately between 70 and 89.
Explanation:
To find the numerical limits for a B grade in a normally distributed set of test scores, we must determine the z-scores that correspond to the top 5% and the bottom 38% of scores since a B grade is assigned to scores below the top 5% and above the bottom 62% (which is equivalent to being below the top 38%). Given the mean score is 73.3 and the standard deviation is 9.7, we can use z-score formulas to convert these percentages to raw scores.
For the top 5% cutoff (z = 1.645), the score is calculated as follows:
Top 5% cutoff score = mean + (z x standard deviation)
Top 5% cutoff score = 73.3 + (1.645 x 9.7)
Top 5% cutoff score = 73.3 + 15.9565
Top 5% cutoff score ≈ 89
For the top 38% cutoff (z = -0.31), which is the bottom 62%, the score is calculated as:
Bottom 62% cutoff score = mean + (z x standard deviation)
Bottom 62% cutoff score = 73.3 + (-0.31 x 9.7)
Bottom 62% cutoff score = 73.3 + (-2.997)
Bottom 62% cutoff score ≈ 70
Therefore, a B grade on this test would be for scores roughly between 70 and 89.
Which of the following is a quantitative continuous variable? a. Where a family goes on vacation b. Distance a family travels on vacation c. Number of vacations a family takes per year d. Number of consecutive years that a family goes on a vacation (Let a trip of at least 200 miles from home be defined as a "vacation") e. Amount spent on the vacation, to the nearest $100.
Answer:
Which of the following is a quantitative continuous variable?
b. Distance a family travels on vacation
e. Amount spent on the vacation, to the nearest $100.
Step-by-step explanation:
Quantitative variables use numbers to express attributes, there are 2 types:
1. Quantitave continuous variables use numbers to express the attributes but, could have an infinite value between its maximum and minimum values, and are able to be divided into smaller values, such as in options b and e.
2. Quantitative discrete, also use numbers to express attributes, but they may have some, but not all the values, and must be integers, such as in options:
c. Number of vacations a family takes per year
d. Number of consecutive years that a family goes on a vacation (Let a trip of at least 200 miles from home be defined as "vacation").
e. Amount spent on the vacation, to the nearest $100.
And there are the cualitative variables, which use words to express attributes, such as in:
a. Where a family goes on vacation.
Before a strike prematurely ended the 1994 major league baseball season, Tony Gwynn of the San Diego Padres had 165 hits in 419 at bats, for a .394 batting average. Baseball fans argued over whether Gwynn was effectively a .400 hitter that year. Test this hypothesis
Answer:
yes, he was effectively a .400 hitter
Step-by-step explanation:
The mean rate of success ( ÿ ) is 0.394.
The total number of hits was 419.
Under table t the value at 5% confidence interval is 1.96.
Here se the asymptotic standard deviation is given as follows
se = [tex]\sqrt{\frac{y(1 - y)}{n} }[/tex]
= [tex]\sqrt{\frac{0.394(1 - 0.394)}{419} }[/tex]
= 0.02387
The confidence interval is calculated as follows:
y ± 1.96 * se
0.394 ± 1.96 * 0.02387
the answer is -0.347 or 0.4407)
meaning that the mean lies between -0.347 to 0.4407.
There is strong evidence that 0.4 lies between the above confidence interval.
This means that Gwynn was effectively a 0.4 hitter
Tony Gwynn's batting average was .394 with 165 hits in 419 at bats. He would have needed approximately 3 more hits to officially reach a .400 average. Despite his impressive performance, it is factually incorrect to label him a .400 hitter for that season.
To test the hypothesis that Tony Gwynn was a .400 hitter in 1994 before the strike ended the season, we need to see if his actual batting average could be considered close to .400 even though it was .394.
Gwynn had 165 hits in 419 at bats, which gives a batting average of 165/419, or approximately .394 when rounded to three decimal places. To have a .400 average, he would need to have 419 * .400 hits, which equals 167.6. Therefore, he needed approximately 3 more hits (rounding up from 2.6) to achieve a .400 average with the same number of at bats.
Given that batting averages are not rounded up after the season ends, stating that Gwynn was a .400 hitter would be incorrect. However, his achievement of .394 is remarkably high, and some may argue that rounding up for discussion purposes could informally suggest he was 'effectively' a .400 hitter.