A survey found that? women\'s heights are normally distributed with mean 63.7 in

ID: 2948988 • Letter: A

Question

A survey found that? women's heights are normally distributed with mean

63.7 in

and standard deviation

2.2 in

The survey also found that? men's heights are normally distributed with mean

67.5 in

and standard deviation

3.9 in

Consider an executive jet that seats six with a doorway height of

55.7 in

Complete parts? (a) through? (c) below.

a. What percentage of adult men can fit through the door without? bending?

The percentage of men who can fit without bending is ?%.

Can you tell me how to enter this into my TI-83 calculator as simple and quick as possible.

Given that,

A survey found that? women's heights are normally distributed with mean

63.7 in

and standard deviation

2.2 in

The survey also found that? men's heights are normally distributed with mean

67.5 in

and standard deviation

3.9 in

Let X be the random variable that mens height.

Here we have to find P(X < 55.7)

Convert x = 55.7 into z-score

z = (x - mean) /sd = (55.7 - 67.5) / 3.9 = -3.03

Now we have to find P(Z < -3.03)

P(Z < -3.03) = 0.0012

TI-83 syntax :

normalcdf (lower bound, upper bound, mean, standard deviation)

lower bound = -99999

upper bound = -3.03

mean = 67.5

sd = 3.9

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