Question

The average American adult has completed an average of 11.25 years of education with a standard deviation of 1.75 years. A random sample of 75 adults is taken.

a) What is the probability that the sample will have a mean more than 11.5?
b) What is the probability that the sample will have a mean between 11 and 12 years?

Answer 1

@ZeHanz

Answer 2

Assume we have a normal distribution.
Because we take the mean value of a sample of 75, the standard deviation is 1.75/√75.
So we have to find the probability: P(X>11.5|\[P(X > 11.5|\mu =11.25,\sigma=1.75/√75)\]

Do you use tables, of a graphical calculator such as the TI-84 for this?

Answer 3

I dont have a calc on me right now

Answer 4

I don't have a table ;)

My TI-84 says:\[NormalCdf(11.5,E99,11,25,1.75/√75))\approx 0.1080\]

Answer 5

Ok so that would be the answer for (a) then?

Answer 6

@ZeHanz

Answer 7

Yes, that would be a.

Answer 8

Ok, how do i do b?

Answer 9

and r u positive a is ;

NormalCdf(11.5, E99, 11, 25, 1.75/sqrt75)), .1080?

Answer 10

b) It's the same sample, so the same standard deviation.
In the calculator function, "NormalCdf", the first two numbers are the left and right boundaries of what you want, so if I replace these numbers with 11 and 12, I get:

NormalCdf(11,12,11.5, 1.75/sqrt(75)) = 0.8919

In my answer above, there is "11,25" in the NormalCdf function. It should be "11.25".
(That's because I'm used to a decimal comma instead of a decimal point).

Answer 11

So a is :
*calc*
NormalCdf(11.5, E99, 11, 25, 1.75/sqrt75)) = .1080

and b is NormalCdf(11,12,11.5, 1.75/sqrt(75)) = 0.8919 ??

Answer 12

Yes, but in the first one, replace "11,25" with "11.25"

Answer 13

Oh ok.. r u sure about these? thanks

Answer 14

and what does the E99 mean?

Answer 15

So this is right? :

(a)
*calc*
NormalCdf(11.5, E99, 11.25, 1.75/sqrt75))
= 0.1080 <---

(b)
*calc*
NormalCdf(11, 12, 11.5, 1.75/sqrt(75))
= 0.8919 <---

Answer 16

E99 is used for infinity. If it has to be greater than 11.5, there is no right bound, so for the function to work, you just put in a large number. "E99" =10^99, so rather large ;)
For that matter, if you would use 1000 instead of E99, this would give the same answer.

This is because the standard deviation is only about 0.2. As you might know, the more standard deviations you are from the mean, the smaller the probability gets.
With a right bound of 1000, you are already 5000 standard deviations away from the mean, so there is nothing there...

What makes you think the answers are wrong?

Answer 17

(a)
*calc*
NormalCdf(11.5, E99, 11.25, 1.75/sqrt75))
= 0.1080 <---

(b)
*calc*
NormalCdf(11, 12, 11.5, 1.75/sqrt(75))
= 0.8919 <---

Answer 18

IS that written right?

Answer 19

@ZeHanz

Answer 20

Right!

Answer 21

thanks can u help me with this?

Answer 22

Describe the sampling distribution of x-bar for samples of size 20 drawn from a normal population with a mean of 50 and standard deviation of 4.

Answer 23

Sorry, I have to go now; maybe later?

Answer 24

I have to get it done now:/

Answer 25

@ZeHanz ?