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. (round to 4 decimal places)

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 and standard deviation , the zscore of a measure X is given by:

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 and standard deviation , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation .

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:

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.

By the Central Limit Theorem

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.