Question

The time it takes to train a newly hired telephone sales representative is normally distributed with a mean of 136 hours and a standard deviation of 8.9 hours. What is the probability that training for a given individual will last between 130 and 140 hours?
Would a good starting point for this question be: P(130 11 years ago

Answer 1

yes, now we could do a change of variable if need be

Answer 2

are you able to use a ti83 or similar?

Answer 3

yes

Answer 4

then all we need is a function called normalcdf

Answer 5

normalcdf(low, high, mean, sd)

Answer 6

2nd, vars, pick the normalcdf function and input the values

Answer 7

how do you input the 140?

Answer 8

And if you will have show work on test i will have to work z score...

Answer 9

are you allowed to use a ti83 on a test?

Answer 10

on my tests, we simply defined the function we used:

normalcdf(130,140,136,8.9) = .4233

Answer 11

test you can use any calculator, but the thing is the teacher expected work wrote out in steps...like calculating the z score, drawing the bell and writing the information in the bell...

Answer 12

if we do a change of variable, then x = mean + z(sd)

Answer 13

P(130<x<140)

P(130 <mean + z(sd) < 140)

P(130-mean < z(sd) < 140-mean)

P((130-mean)/sd < z < (140-mean)/sd )

Answer 14

I've never actually seen this you are showing me, so what the have us do is z=x-u/o...and i was thinking that you would find both z scores and subtract them to get the answer???

Answer 15

subtracting z scores does nothing but tell us how many z scores fit between the interval. but we dont have a uniform distribution so 3 sd between a and b, is going to have a different value than 3 sd between c and d

Answer 16

z scores allow us to look up tables that tell us the amount of area (probability) associted with it

Answer 17

its those areas that we will work with

Answer 18

but then tables are formatted by author preferences, so how we use a table will depend on your authors format

Answer 19

Yes that is very true. But i see what you are showing me with the calculator but the study answers shows 0.4222 as the answer. Confused at moment

Answer 20

.4233 is more exact, so it depends on where in the process you start rounding things and what you use to reference them with

Answer 21

how did you enter the 130 and 140 values in calculator together?

Answer 22

130-136 = 6/8.9 = 0.6741... = -0.67

140-136 = 4/8.9 = 0.4494... = 0.45

in the ti83 we have to get to the distribution menu: 2nd, VARS, then choose the normalcdf

that prompts you with: normalcdf(

input: 130, 140, 136, 8.9) and hit enter

Answer 23

the normalcdf function takes 4 parameters: low, high, mean, sd

Answer 24

http://www.wolframalpha.com/input/?i=normalcdf%28-.67%2C.45%2C0%2C1%29

notice that when we rnd the z scores to 2 decies, we get the .4222 solution

Answer 25

yes I got z scores of-.067 for 130 and 0.44 for 140

Answer 26

the calculator wont rnd them, they will use them more precisely, therefore the 'difference' in solutions

Answer 27

so i looked at the link you send and i plugged in the my numbers and I'm seeing what you are saying, but the z score of 130=-.67 which on table is 0.2514 and 140=.44 which on table is 0.6700. So what do you do with all that info to get 0.4222/0.4233?

Answer 28

so you have add the two scores together and then subtract 1 from sum of z scores

Answer 29

your z scores are giving you left tail areas

Answer 30

b is to big, a is to small, the middle part of it is just b-a

Answer 31

and your z for 140 should rnd to .45

Answer 32

Ok now I believe i have it. 130=-0.67=0.2514 and 140=0.45=0.6736. So 0.6736-0.2514=0.4222

Answer 33

seems reasonable to me

Answer 34

Thanks for your help and patience i feel a little slow after this;)

Answer 35

yw, good luck :)