Question

# Q1 Part A.) A survey found that​ women's heights are normally distributed with mean 63.1 in....

Q1

Part A.)

A survey found that​ women's heights are normally distributed with mean 63.1 in. and standard deviation 3.5 in. The survey also found that​ men's heights are normally distributed with mean 68.6 in. and standard deviation 3.2 in. Consider an executive jet that seats six with a doorway height of 55.8 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
​%.
​(Round to two decimal places as​ needed.)

Part B.)

Physicians want to select a minimum temperature for requiring further medical tests. What should that temperature​ be, if we want only​ 5.0% of healthy people to exceed​ it? (Such a result is a false​ positive, meaning that the test result is​ positive, but the subject is not really​ sick.)
. The minimum temperature for requiring further medical tests should be degrees Upper F if we want only​ 5.0% of healthy people to exceed it.

Part A)

As the data is normally distributed, we can use standard normal z table to estimate the answer

For men

Mean = 68.6

S.d = 3.2

Now height is 55.8

So all the men with height less than 55.8 can enter without bending

So, we need to find p(x<55.8)

Z = (55.8-68.6)/3.2

Z = -4

From z table, p(z<4) = 0.000032

That is 0.000032*100%

= 0.0032%

= 0%

In part 2

Data is missing

You have.not provided the mean and standard deviation

But i can show you the process

From z table, p(z>1.645) = 0.05

So, z = 1.645

And z is given by (x-mean)/s.d

Now you need to determine x here

1.645 = (x-mean)/standard deviation

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