For the most recent year available, the mean annual cost to attend a private university in the United States was $20,232. Assume the distribution of annual costs follows the normal probability distribution and the standard deviation is $4,350.

Ninety-five percent of all students at private universities pay less than what amount? (Round z value to 2 decimal places and your final answer to the nearest whole number

Ninety-five percent of all students at private universities pay less than $27,366.

Step-by-step explanation:

Problems of normally distributed samples can be 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.

In this problem, we have that:

Ninety-five percent of all students at private universities pay less than what amount?

They will pay less than the 95th percentile, that is, less than X when Z = 1.64. So

Ninety-five percent of all students at private universities pay less than $27,366.

Step-by-step explanation:

Since the cost to attend a private university in the United States, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = cost to attend a private university.

µ = mean cost

σ = standard deviation

From the information given,

µ = $20232

σ = 4350

The probability of Ninety-five percent of all students at private universities is 0.95. Looking at the normal distribution table, the z is score corresponding to 0.95 is 1.65

Therefore

1.645 = (x - 20232)/4350

Cross multiplying, it becomes

1.65 × 4350 = x - 20232

x = 7177.5 + 20232 = 27409.5

x = 27410 to the nearest whole number

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