In a survey of men aged 20-29 in a certain country, the mean height is 73.4 inches with a standard deviation of 2.7 inches. Find the minimum height in the top 15% of heights. Use a TI-83, TI-83 plus, or TI-84 calculator, and round your answer to one decimal place.

The minimum height in the top 15% of heights is 76.2 inches.

Step-by-step explanation:

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.

In this problem, we have that:

Find the minimum height in the top 15% of heights.

This is the value of X when Z has a pvalue of 0.85. So it is X when Z = 1.04.

The minimum height in the top 15% of heights is 76.2 inches.

5/6

step-by-step explanation:

if i remember correctly with perpendicular stuff you just take away the negative

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x