A random sample of items is selected from a population of size . What is the probability that the sample mean will exceed if the population mean is and the population standard deviation eq

The probability is 1 subtracted by the p-value of , in which X is the value we want to find the probability of the sample mean exceeding, is the population mean, is the standard deviation of the population and n is the size of the sample.

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 normal distributions can be solved using the z-score formula.

In a set with mean and standard deviation , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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.

Central Limit Theorem for the sample mean:

Sample of size n, and thus:

Probability of the sample mean exceeding a value:

The probability is 1 subtracted by the p-value of , in which X is the value we want to find the probability of the sample mean exceeding, is the population mean, is the standard deviation of the population and n is the size of the sample.

answer:

125 yards

explanation:

75+50=125. that is my answer