Question

Models in the United States and England work very hard to keep their weight down. Many participate in weightloss programs, carefully monitor their diet, and exercise regularly. The weight of a model is approximately normal with mean 52 kg and standard deviation 1.2 kg. Suppose a model is randomly selected…

3 parts inside, pls explain? thanks!:)

Answer 1

1) what is the probability that the model weighs more than 53 kg? (round to 4 decimal places)
2)what is the probability that the model weighs between 50 and 54 kg? (round to 3 decimal places)
3) A value w such that 80% of all models weigh more than w is.. (round to 2 decimal places)

Thanks:)

Answer 2

:) one minute

Answer 3

thank you! i forget the process and steps for how to come to the solutions! much appreciated:)

Answer 4

mean 52kg
standard deviation 1.2kg

can you draw the normal curve first ?

Answer 5

? not sure how to label :(

Answer 6

@ganeshie8 ? :)

Answer 7

that looks good, just label the mean at center

Answer 8

like so?

Answer 9

Yes, now look at first problem

Answer 10

1) what is the probability that the model weighs more than 53 kg? (round to 4 decimal places)

Answer 11

do you know how to find the zscore ?

Answer 12

ohhh ok! and a bit confused about z score; how do i calculate that?

Answer 13

observation = 53
mean = 52
standard deviation = 1.2

\[\text{zscore} = \dfrac{\text{observation - mean}}{\text{standard deviation}} \]

Answer 14

ohh okay! so i calculate the z score?

and i get
zscore = 53-52/1.2 = 1/1.2 = 0.8333 and so the answer for 1 is the probability is 83.3333 percent?

Answer 15

Nope, zscore is not probability.

Answer 16

ooh okie, so what happens next?:)

Answer 17

For probability, you need to look up ztable for a zscore of 0.8333

Answer 18

do you have a ztable ?

Answer 19

oh okay! yes:) so i look up and it is closest to 0.8340 so the probability is 7 percent? is that correct?

Answer 20

how did u get 7 percent ?

Answer 21

my table says the area corresponding to the zscore of 0.833 is .7967

Answer 22

oh oops yes, i looked at the wrong area sorry about that!! :(

okay! so then is it i go 1-0.7967?

Answer 23

My table gives the area to the left of 0.833

But we're looking for the right side area, so you subtract 0.7967 from 1 yeah :)

Answer 24

for part a, do we get the probability as 20.23 % ?

Answer 25

okie!! so 1-0.7967 = 0.2033 and that is the solution for part 1?

Answer 26

Looks good!

Answer 27

Oh well it looks like you have this @ganeshie8 :) so i guess you dont need my help?

Answer 28

@ganeshie8 connection's back!! how do you do part 2?

Answer 29

2)what is the probability that the model weighs between 50 and 54 kg? (round to 3 decimal places)

Answer 30

start by finding the corresponding zscores for the observations 50 and 54

Answer 31

okay! so i do this?

observation 50 zscore = 50-52/1.2 = -2/1.2 = -1.666666667
observation 54 z score = 54-52/1.2 = 2/1.2 = 1.666666667

?

Answer 32

Yes, look up the ztable and find the corresponding areas too

Answer 33

okay, so we get 0.1423 & 0.8577 ?

Answer 34

Both are wrong, try again

Answer 35

you may use wolfram to double check
http://www.wolframalpha.com/input/?i=P%28Z%3C++%2850-52%29%2F1.2%29

Answer 36

oh! so 0.0446 & 0.9554?

Answer 37

oops sorry, 0.95221 & 0.0477904

Answer 38

are you looking at -1.7 ?

that is too much rounding, please look at -1.67 maybe

Answer 39

`oops sorry, 0.95221 & 0.0477904`


these look much better :)

Answer 40

and so when i subtract and round to three digits, do i get 0.905?

Answer 41

im getting 0.904 ?

Answer 42

wolfram also says it is 0.904

http://www.wolframalpha.com/input/?i=P%28%2850-52%29%2F1.2+%3C+Z+%3C+%2854-52%29%2F1.2%29

Answer 43

ooh okie!! thank you!! how do we do part 3 now? sorry, os keeps kicking me off:(

Answer 44

3) A value w such that 80% of all models weigh more than w is.. (round to 2 decimal places)

Answer 45

can you read part 3 and tell me what exactly we are after here ?

Answer 46

we are after something like this?