At a large university, the mean amount spent by students for cell phone service is $58.90 per month with a standard deviation of $3.64 per month. Consider a group of 44 randomly chosen university students. What is the probability that the mean amount of their monthly cell phone bills is more than $60?

2.28% probability that the mean amount of their monthly cell phone bills is more than $60

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 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 problem, we have that:

What is the probability that the mean amount of their monthly cell phone bills is more than $60?

This is 1 subtracted by the pvalue of Z when X = 60. So

By the Central Limit Theorem

has a pvalue of 0.9772

1 - 0.9772 = 0.0228

2.28% probability that the mean amount of their monthly cell phone bills is more than $60

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step-by-step explanation: