Answer:
3.24% probability that the mean of the sample would differ from the population mean by greater than 30 points
Explanation:
To solve this question, we have 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 random variable X, with mean
and standard deviation
, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation, which is also called standard error
In this problem, we have that:
![\mu = 1612, \sigma = 260, n = 344, s = (260)/(√(344)) = 14.02[qtex]</p><p><strong>What is the probability that the mean of the sample would differ from the population mean by greater than 30 points</strong></p><p>Either it differs by 30 or less points, or it differs by more than 30 points.</p><p>The sum of the probabilities of these events is 100. So i will first find the probability that it differs by 30 or less points.</p><p><strong>Probability that it differs by 30 or less points</strong></p><p>Pvalue of Z when X = 1612 + 30 = 1642 subtracted by the pvalue of Z when X = 1612 - 30 = 1582. So</p><p><strong>X = 1642</strong></p><p>[tex]Z = (X - \mu)/(\sigma)]()
By the Central Limit Theorem
has a pvalue of 0.9838
X = 1582
has a pvalue of 0.0162
0.9838 - 0.0162 = 0.9676
96.76% probability that it would differ by 30 or less points.
What is the probability that the mean of the sample would differ from the population mean by greater than 30 points
96.76 + p = 100
p = 3.24
3.24% probability that the mean of the sample would differ from the population mean by greater than 30 points