Question

In a survey of men aged 20-29 in a certain country, the mean height is 73.4...

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.
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Answer #1

Solution:

Given: Men aged 20-29 in a certain country, the mean height is 73.4 inches with a standard deviation of 2.7 inches.

Lets assume X follows a Normal distribution with mean= \mu = 73.4 inches and a standard deviation \sigma =   2.7 inches.

We have to find :  the minimum height in the top 15% of heights.

That is find x value such that:

P( X > x ) =15%

P( X > x ) =0.1500

thus we can write this as:

P( X < x ) =1 - P( X > x)

P( X < x ) =1 - 0.1500

P( X < x ) = 0.8500

That means: if area above x is 15% then area below x is 85%

Now use following steps in TI 84 plus calculator:

1) Press 2ND and press VARS

2) Select invNorma(

TI-84 Pius Silver Edition TEXAS INSTRUMENTS DISTR DRAW 1: normalpdf 2: normalcdfc invNorma 4. iFT 5: tpdf 6:tcdf 7*X2 pdf STA

Enter Numbers:

for area we use left area , so we enter 0.8500 area.

Mean = \mu = 73.4

Standard deviation = \sigma = 2.7

TI-84 Plus Silver Edition TEXAS INSTRUMENTS invnom area: 6.8500 ONNU N- STAT PLOTF1 TBLSETE FORMAT E WINDOW CALC F4 TRACE TAB

Click on Paste and press ENTER two times

TI-84 Plus Silver Edition TEXAS INSTRUMENTS invNorm(0.8500, 76.19837013 STAT PLOTF1 TBLSETF2 FORMAT ES CALC F4 WINDOW ZOON TR

Thus we get:

x = 76.19837

x = 76.2

Thus the minimum height in the top 15% of heights is x = 76.2 inches.

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