Question

PLEASE HELP!!! WILL GIVE MEDALS FAST!!! A survey found that women’s heights are normally distributed with mean 63.8 in and a standard deviation 2.3 in. A branch of the military requires women’s heights to be between 58 in and 80 in.
a. Find the percentage of women meeting the height requirement. Are many women being denied the opportunity to join this branch of the military because they are too short or too tall?
b. If this branch of the military changes the height requirements so that all women are eligible except the shortest 1% and the tallest 2%, what are the new height requirements.

Answer 1

so do you use your z-table for this?

Answer 2

Yes and I must show my work no idea where to start

Answer 3

do you know how to use the z-table, ...
.... to find what proportion of women have height less than 58in. ?

Answer 4

Do I use the negative z score table?

Answer 5

to get the z value you use the formula:\[\Large z= {x-\mu \over \sigma}\]

Answer 6

Okay...

Answer 7

when x=58
\[z={58-63.8 \over 2.3}\approx -2.52\]

Answer 8

Okay

Answer 9

so whats the probability at z=-2.52

Answer 10

? what did you get in the z-table?

Answer 11

.0059

Answer 12

right ! .. so 0.59 % have height less than 58in

Answer 13

do the same thing to find the % of women with height less than 80in.

Answer 14

Ok

Answer 15

7.0

Answer 16

right .. and you won't even find that on your z-table ... lol

Answer 17

so that means ... ?

Answer 18

that pretty muchh 100% women have height below 80 in.

Answer 19

so (100 - 0.59)% = 99.41% have heights between 58 and 80.

Answer 20

Okay...

Answer 21

does it make sense ... or are you still kinda lost?

Answer 22

So the answer to a. Is 99.4 and b. Is 58 and 80?

Answer 23

we just did part a.

you to start fresh for part b.

Answer 24

Oh ok I got it so far

Answer 25

for the shortest 1% find the z-value that has probability closest to 0.01

Answer 26

Okay...

Answer 27

what do you get from the z-table?

Answer 28

-2.33 = .0099

Answer 29

right ... now convert z=-2.33 to x to get the height it represents

Answer 30

.02277

Answer 31

?

Answer 32

.002277 + 63.8= 63.82?

Answer 33

\[z= {x-\mu \over \sigma}\]

\[\implies x= z \cdot \sigma+\mu\]

Answer 34

-2.33*2.3+63.8

Answer 35

Sorry I'm so lost :) the answer is 58.4

Answer 36

\[\checkmark\]the same way for top 2% ... find the z where probability is closest to 0.98

Answer 37

-2.34 = .0096

Answer 38

tallest 2% means 98% are below that height ...

z=0 has 0.5000 or 50% .. more than 50% is positive z

Answer 39

Ok...

Answer 40

closest to 0.9800

Answer 41

-2.34 is what I've found

Answer 42

not 0.0098

it's supposed to be 0.9800

Answer 43

look on the positive side...

Answer 44

Okay...

Answer 45

2.05 = .9798

Answer 46

right :)

Answer 47

convert this z=2.05 to height ... and you have your answer.

Answer 48

so what is it?

Answer 49

68.5%

Answer 50

68.5 inches :p

Answer 51

So that's it?

Answer 52

yep ... the new height requirement is you must be between 58.4 and 68.5 inches

Answer 53

Okay thanks