Answer:
0.4122 = 41.22% probability that the sample mean would differ from the true mean by greater than 1 dollar
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(which is the square root of the variance)
, 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 random variable X, with mean and standard deviation
, the sample means with size n of at least 30 can be approximated to a normal distribution with mean
and standard deviation
In this problem, we have that:
What is the probability that the sample mean would differ from the true mean by greater than 1 dollar?
Either it differs by 1 dollar or less, or it differs by more than one dollar. The sum of the probabilities of these events is decimal 1.
Probability it differs by 1 dollar or less:
pvalue of Z when X = 50+1 = 51 subtracted by the pvalue of Z when X = 50 - 1 = 49.
X = 51
By the Central Limit Theorem
has a pvalue of 0.7939
X = 49
has a pvalue of 0.2061
0.7939 - 0.2061 = 0.5878
Probability it differs by more than 1 dollar:
p + 0.5878 = 1
p = 0.4122
0.4122 = 41.22% probability that the sample mean would differ from the true mean by greater than 1 dollar