Question

Full-time college students report spending a mean of 29 hours per week on academic activities, both inside and outside the classroom. Assume the standard deviation of time spent on academic activities is 5 hours. Complete parts (a) through (d) below.

a. If you select a random sample of 25 full-time college students, what is the probability that the mean time spent on academic activities is at least 28 hours per week? ___(Round to four decimal places as needed.)

b. If you select a random sample of 25 full-time college students, there is an 84 % chance that the sample mean is less than how many hours per week? ___ (Round to two decimal places as needed.)

c. What assumption must you make in order to solve (a) and (b)? (choose between A through D)

A. The population is symmetrically distributed, such that the Central Limit Theorem will likely hold for samples of size 25.

B. The sample is symmetrically distributed, such that the Central Limit Theorem will likely hold.

C. The population is uniformly distributed.

D. The population is normally distributed.

d. If you select a random sample of 64 full-time college students, there is an 84 % chance that the sample mean is less than how many hours per week? ___(Round to two decimal places as needed.)

Answer 1

Answer:

a) 0.8413

b) 30

c) A. The population is symmetrically distributed, such that the Central Limit Theorem will likely hold for samples of size 25.

d) 29.63

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 sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 29, \sigma = 5$

a. If you select a random sample of 25 full-time college students, what is the probability that the mean time spent on academic activities is at least 28 hours per week?

25 students, so $n = 25, s = \frac{5}{\sqrt{25}} = 1$

This is 1 subtracted by the pvalue of Z when X = 28. So

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

By the Central Limit Theorem

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

$Z = \frac{28 - 29}{1}$

$Z = -1$

$Z = -1$ has a pvalue of 0.1587

1 - 0.1587 = 0.8413

0.8413 is the answer.

b. If you select a random sample of 25 full-time college students, there is an 84 % chance that the sample mean is less than how many hours per week?

Value of X when Z has a pvalue of 0.84. So X when Z = 1.

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

$1 = \frac{X - 29}{1}$

$X - 29 = 1$

$X = 30$

c. What assumption must you make in order to solve (a) and (b)? (choose between A through D)

Central limit theorem works if the population is normally distributed, or if the sample means are of size at least 30

So the correct answer is:

A. The population is symmetrically distributed, such that the Central Limit Theorem will likely hold for samples of size 25.

d. If you select a random sample of 64 full-time college students, there is an 84 % chance that the sample mean is less than how many hours per week?

Now we have n = 64, so $s = \frac{5}{\sqrt{64}} = 0.63$

Value of X when Z has a pvalue of 0.84. So X when Z = 1.

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

$0.63 = \frac{X - 29}{1}$

$X - 29 = 0.63$

$X = 29.63$

Answer 2

This question is based on the normal probability and central limit theorem.Therefore, we conclude the answer, (a) 0.8413, (b) X = 30, (c) A and (d) X = 29.63.

Given:

Full-time college students report spending a mean of 29 hours per week on academic activities, both inside and outside the classroom. Assume the standard deviation of time spent on academic activities is 5 hours.

According to this question,

Firstly, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:  Problems of normally distributed samples are solved by  using the z-score formula.

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

$Z = \dfrac{\overline{X}-\mu}{\sigma}$

Central limit theorem:  This states that, for a normally distributed random variable X, with mean  $\mu$ and standard deviation $\sigma$ , the sample mean with size n can be approximated to a normal distribution with mean  and standard deviation :

$s = \dfrac{\sigma}{\sqrt{n} }$

In this problem, it is given that:

$\mu = 29 , \sigma = 5$

(a) In this we have to find the probability that the mean time spent on academic activities is at least 28 hours per week,

$s = \dfrac{\sigma}{\sqrt{n} } = \dfrac{5}{\sqrt{25} } = 1$

Thus, 1 subtracted from the p-value of Z when X = 28. So,

By central limit theorem,

$Z = \dfrac{\overline{X}-\mu}{\sigma} = \dfrac{28-29}{1} = -1$ has a p-value of 0.1587.

Thus, 1 - 0.1587 = 0.8413

Hence, the probability that the mean time spent on academic activities is at least 28 hours per week is 0.8413.

(b) If you select a random sample of 25 full-time college students, what is the probability that the mean time spent on academic activities is at least 28 hours per week then,

Value of X, when Z has a p-value of 0.84. So, X, when Z = 1.

$Z = \dfrac{\overline{X}-\mu}{\sigma} \\\\1= \dfrac{\overline{X}-29}{1} \\\\\overline{X}-29 = 1\\\\\overline{X} = 30$

(c) We have to assumed that, central limit theorem works at , if the population is normally distributed, or if the sample means are of size at least 30.

Thus, the correct answer is:  (A) The population is symmetrically distributed, such that the central Limit Theorem will likely hold for samples of size 25.

(d) If you select a random sample of 64 full-time college students, there is an 84 % chance that the sample mean is less than how many hours per week then,

$s = \dfrac{\sigma}{\sqrt{n} } = \dfrac{5}{\sqrt{64} } = 0.63$

Value of X, when Z has a p-value of 0.84. So, X when Z = 1,

$Z = \dfrac{\overline{X}-\mu}{\sigma} \\\\0.63= \dfrac{\overline{X}-29}{1} \\\\\overline{X}-29 = 0.63\\\\\overline{X} = 29.63$

Therefore, we conclude the answer, (a) 0.8413, (b) X = 30, (c) A and (d) X = 29.63.

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