A normal population has a mean of 58 and a standard deviation of 13. You select a random sample of 25. Compute the probability the sample mean is: (Round your z values to 2 decimal places and final answers to 4 decimal places): Greater than 60.

0.2206 = 22.06% probability that the sample mean is greater than 60.

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 and standard deviation

In this problem, we have that:

Compute the probability the sample mean is greater than 60:

This probability is 1 subtracted by the pvalue of Z when X = 60. So

By the Central Limit Theorem

has a pvalue of 0.7794

1 - 0.7794 = 0.2206

0.2206 = 22.06% probability that the sample mean is greater than 60.

0.0076, if you meant 4.125 divided by 0.03125 it would be 132

step-by-step explanation:

first you have to move the decimal in 0.03125 over until its behind the 5 which is 5 places over. now the number is 3125

then you have to move the decimal in 4.125, 5 places over. now the number is 4,112,500.

now you divide like you would normally would and get 0.0076

step-by-step explanation:

gym b charges $1.05 less per hour than gym a.