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pentagon [3]
3 years ago
5

ASK YOUR TEACHER Time taken by a randomly selected applicant for a mortgage to fill out a certain form has a normal distribution

with mean value 8 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
Mathematics
1 answer:
Arlecino [84]3 years ago
5 0

Answer:

On the first day, 98.75% probability that the sample average amount of time taken is at most 11 min.

On the second day, 99.29% probability that the sample average amount of time taken 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 normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes 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.

In this problem, we have that:

\mu = 8, \sigma = 3

First day

n = 5, s = \frac{3}{\sqrt{5}} = 1.34

The probability is the pvalue of Z when X = 11. So

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

By the Central Limit Theorem

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

Z = \frac{11 - 8}{1.34}

Z = 2.24

Z = 2.24 has a pvalue of 0.9875

On the first day, 98.75% probability that the sample average amount of time taken is at most 11 min.

Second day

n = 6, s = \frac{3}{\sqrt{5}} = 1.2247

The probability is the pvalue of Z when X = 11. So

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

By the Central Limit Theorem

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

Z = \frac{11 - 8}{1.2247}

Z = 2.45

Z = 2.45 has a pvalue of 0.9929

On the second day, 99.29% probability that the sample average amount of time taken is at most 11 min.

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