The standard deviation of the scores on a skill evaluation test is 260 points with a mean of 1612 points. If 344 tests are sampled, what is the probability that the mean of the sample would differ from the population mean by greater than 30 points

3.24% probability that the mean of the sample would differ from the population mean by greater than 30 points

Step-by-step 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:

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

ab will be equal to

answer: 10 meter

step-by-step explanation:

need details