Thanks for the help! Been stuck on this one for a little while.
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Answer:
96.6% probability that the mean of a sample would differ from the population mean by less than 126 miles
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
A reminder is that the standard deviation is the square root of the variance.
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:
Probability that the mean of the sample would differ from the population mean by less than 126 miles
This is the pvalue of Z when X = 3639 + 126 = 3765 subtracted by the pvalue of Z when X = 3639 - 126 = 3513. So
X = 3765
By the Central Limit Theorem
has a pvalue of 0.983
X = 3513
has a pvalue of 0.017
0.983 - 0.017 = 0.966
96.6% probability that the mean of a sample would differ from the population mean by less than 126 miles