The time for a visitor to read health instructions on a Web site is approximately normally distributed with a mean of 10 minutes

Question

18 days ago

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The time for a visitor to read health instructions on a Web site is approximately normally distributed with a mean of 10 minutes

and a standard deviation of 2 minutes. Suppose 64 visitors independently view the site. Determine the following: a. The expected value and the variance of the mean time of the visitors. b. The probability that the mean time of the visitors is within 15 seconds of 10 minutes c. The value exceeded by the mean time of the visitors with probability 0.01.

Mathematics

1 answer:

Svet_ta [12.7K] · Answer 1 · 2026-04-12T00:36:48+03:00

Response:

a) The average is 10 and the variance is 0.0625.

b) 0.6826 = 68.26% likelihood that the average time of visitors falls within 15 seconds of 10 minutes.

c) 10.58 minutes.

Step-by-step clarification:

To figure this out, we must comprehend the normal probability distribution alongside the central limit theorem.

Normal Probability Distribution

Issues involving normal distributions can be resolved using the z-score formula.

For a data set with mean $\mu$ and standard deviation $\sigma$ , the z-score related to a measure X is defined as:

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

The z-score illustrates how many standard deviations the measure is positioned from the mean. Once the z-score is calculated, we refer to the z-score table to find the p-value matching this z-score. This p-value reflects the likelihood that the measurement is less than X, which means it indicates the percentile of X. To determine the chance that the measure exceeds X, we subtract the p-value from 1.

Central Limit Theorem

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

For non-normal distributions, this theorem applies if n is at least 30.

In this case, the distribution has a mean of 10 minutes and a standard deviation of 2 minutes.

This indicates that $\mu = 10, \sigma = 2$

Assuming 64 visitors independently access the site.

This implies that $n = 64, = \frac{2}{\sqrt{64}} = 0.25$

a. The expected value and variance of the average time of the visitors.

Employing the Central Limit Theorem, the mean is 10 and the variance is (0.25)^2 = 0.0625.

b. The chance that the mean visit duration is within 15 seconds of 10 minutes.

15 seconds = 15/60 = 0.25 minutes, thus between 9.75 and 10.25 seconds, which corresponds to the p-value of z when X = 10.25 minus the p-value of z when X = 9.75.

X = 10.25

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

Through the Central Limit Theorem

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