The time a student sleeps per night has a distribution with mean 6.3 hours and a standard deviation of 0.6 hours. Find the probability that average sleeping time for a randomly selected sample of 42 students is more than 6.5 hours per night. Answer: (round to 4 decimal places)

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Answer:

0.0154 = 1.54% probability that average sleeping time for a randomly selected sample of 42 students is more than 6.5 hours per night.

Step-by-step explanation:

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

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

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

In this question, we have that:

[tex]\mu = 6.3, \sigma = 0.6, n = 42, s = \frac{0.6}{\sqrt{42}} = 0.0926[/tex]

Find the probability that average sleeping time for a randomly selected sample of 42 students is more than 6.5 hours per night.

This is 1 subtracted by the pvalue of Z when X = 6.5.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{6.5 - 6.3}{0.0926}[/tex]

[tex]Z = 2.16[/tex]

[tex]Z = 2.16[/tex] has a pvalue of 0.9846

1 - 0.9846 = 0.0154

So

0.0154 = 1.54% probability that average sleeping time for a randomly selected sample of 42 students is more than 6.5 hours per night.