Answer:
Since |Z| = 1.66 < 2, this outcome should not be considered unusual.
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.
If |Z| > 2, X is considered unusual.
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:
A random sample of 60 customers has a mean of 36.1 minutes or less. Would this outcome be considered unusual, so that the store should reconsider its displays?
We have to find Z when X = 36.1.
By the Central Limit Theorem
So |Z| = 1.66
Since |Z| = 1.66 < 2, this outcome should not be considered unusual.