Mathematics, 20.07.2021 02:20 cbender30p860we

Time taken by a randomly selected applicant for a mortgage to fill out a certain form has a normal distribution with mean value 10 min and standard deviation 3 min. If five individuals fill out a form on one day and six on another, what is the probability that the sample average amount of time taken on each day is at most 11 min

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Answer from: mrashrafkotkaat

0.6121 = 61.21% probability that the sample average amount of time taken on each day is at most 11 min.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Normal distribution with mean value 10 min and standard deviation 3 min.

This means that $\mu = 10, \sigma = 3$

First day:

5 individuals, so $n = 5, s = \frac{3}{\sqrt{5}}$

The probability is the p-value of Z when X = 11. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{11 - 10}{\frac{3}{\sqrt{5}}}$

Z = 0.745

Z = 0.745 has a p-value of 0.7719.

Second day:

6 individuals, so $n = 6, s = \frac{3}{\sqrt{6}}$

$Z = \frac{X - \mu}{s}$

$Z = \frac{11 - 10}{\frac{3}{\sqrt{6}}}$

Z = 0.817

Z = 0.817 has a p-value of 0.793.

What is the probability that the sample average amount of time taken on each day is at most 11 min?

Each day is independent of other days, so we multiply the probabilities.

0.7719*0.793 = 0.6121

0.6121 = 61.21% probability that the sample average amount of time taken on each day is at most 11 min.

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Answer from: Quest

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step-by-step explanation:

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