If np >5 and nq>5 estimate P (fewer than 4) with n=13 p=0.4 by using normal distribution as an approximate to the binomial distribution if np <5 or nq< then state that the normal approximation is not suitable.

To estimate the probability of getting fewer than 4 successes in a binomial distribution, we can use the normal distribution as an approximation based on the conditions np > 5 and nq > 5. By calculating the mean and standard deviation of the normal distribution, we can use a z-score to find the estimated probability. In this case, the estimated probability is approximately 0.1709.

The given question requires us to estimate the probability of getting fewer than 4 successes in a binomial distribution. To determine if we can use the normal distribution as an approximation, we need to check if both np and nq are greater than 5. In this case, np = 13 * 0.4 = 5.2 and nq = 13 * 0.6 = 7.8, so both conditions are satisfied. Thus, we can use the normal distribution as an approximation. Let's estimate P(fewer than 4) using the normal distribution.

The mean of the normal distribution is given by μ = np = 13 * 0.4 = 5.2. The standard deviation is calculated using the formula σ = √(npq), where q = 1 - p. So, q = 1 - 0.4 = 0.6. Plugging in the values, we get σ ≈ √(13 * 0.4 * 0.6) ≈ 1.789. Now, we can use the normal distribution to estimate the probability.

Using the z-score formula z = (x - μ) / σ, where x is the number of successes we want to estimate the probability for, we can find the z-score for 3.5 (less than 4). z = (3.5 - 5.2) / 1.789 ≈ -0.951. Now, we can use a standard normal distribution table or calculator to find the cumulative probability for this z-score. P(fewer than 4) = P(z < -0.951) ≈ 0.1709 (approximately). Therefore, the estimated probability of getting fewer than 4 successes in the given binomial distribution is approximately 0.1709.

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Firstly, we would have to evaluate whether a normal approximation would be useful to use. However, given our conditions where `n` is 13, `p` is 0.4, `q` (the complement of `p`) is 1 - 0.4 or 0.6, we can clearly use the normal approximation to the binomial distribution because both `np` (13 * 0.4 = 5.2) and `nq` (13 * 0.6 = 7.8) are greater than 5.

For the normal approximation, we begin by calculating the mean (µ) and standard deviation (σ) of the binomial distribution. The mean of a binomial distribution is calculated as `np` hence in our case, `np` is 13 * 0.4 which is equal to 5.2. To find the standard deviation, we use the formula `sqrt(npq)`, so in our case, we have`sqrt(13 * 0.4 * 0.6)`, which is approximately 1.76635.

Now, to find the probability of getting fewer than 4, we would calculate a z-score for the value of x = 4. This helps convert the raw scores in a dataset into a standardised form, allowing us to better understand the position of individual scores in relation to the rest of the dataset. The z-score is calculated as `z = (x - µ) / σ` . With x being 4, µ is 5.2, and σ is 1.76635, our z-score will be (4 - 5.2) / 1.76635, which gives us approximately -0.67937.

Now, the last step is to find the cumulative probability associated with this z-score. In essence, the cumulative probability associated with a z-score tells you the probability that a certain observation will occur given a standard normal distribution (which is our case, as we've standardised our dataset using the z-score). Checking a z-table, or by using the cumulative distribution function for a value of z = -0.67937, the associated probability is approximately 0.24845.
So the final probability that we will get fewer than 4 is approximately 0.24845 or 24.85% when we use the normal distribution as an approximation to the binomial distribution.

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