S3 help

Need some help with a qs.

The heights of a population of male students are distributed normally with mean 178cm and standard deviation 5cm. The heights of a population of female students are distributed normally with mean 168cm and standard deviation 4cm.
Find the probability that a randomly chosen female is taller than a randomly chosen male.

So I tried subtracting the means and adding the s.d's.

I got F-M - (-10, 9)
Then idk. I tried F-m>0
But that didn't work.

Thanks

Reply 1

Kevin De Bruyne

Original post by Super199

Check the variance of F-M again.

Reply 2

Super199

Original post by SeanFM

Check the variance of F-M again.

4-5?

Surely that would give a really big z value?

0-10/-1

Reply 3

Kevin De Bruyne

Original post by Super199

4-5?

Surely that would give a really big z value?

0-10/-1

Var(aX+bY) = ?

(where X and Y are indep.)

(edited 7 years ago)

Reply 4

Super199

Original post by SeanFM

Var(aX+bY) = ?

(where X and Y are indep.)

a^2var(x)+b^2var(y)

What are the values of a and b?

Is it 16 and 25. idek sorry

Reply 5

Kevin De Bruyne

Original post by Super199

a^2var(x)+b^2var(y)

What are the values of a and b?

Is it 16 and 25. idek sorry

That is correct. In this case (as you have worked out already, to give Var(F-M) = 16 + 25) X = F, a = 1, Y = M, b = -1.

Reply 6

Super199

Original post by SeanFM

That is correct. In this case (as you have worked out already, to give Var(F-M) = 16 + 25) X = F, a = 1, Y = M, b = -1.

Im confused.

a^2var(x) +b^2var(y)

In the qs is a = 4 and b= 5. But whats var(x) and var(y)?

I don't get what you have wrote in bold, because I thought you add the variances anyway.

Reply 7

Kevin De Bruyne

Original post by Super199

Im confused.

a^2var(x) +b^2var(y)

In the qs is a = 4 and b= 5. But whats var(x) and var(y)?

I don't get what you have wrote in bold, because I thought you add the variances anyway.

Forget about the X's and Y's, I think it is too confusing - let's stick with M and F.

Var(M-F) = 1^2 * Var(M) + (-1)^2 * Var(F) = Var(M) + Var(F), and remember that you are given standard deviation rather than variance.

Reply 8

Super199

Original post by SeanFM

P(z> (0-10)/ root 41)?

Reply 9

Kevin De Bruyne

Original post by Super199

P(z> (0-10)/ root 41)?

Correct

Reply 10

Super199

Original post by SeanFM

Correct

That doesn't give me the right answer lol. It gives me 0.94083.
The answer in the book is 0.0592

Reply 11

Kevin De Bruyne

Original post by Super199

That doesn't give me the right answer lol. It gives me 0.94083.
The answer in the book is 0.0592

Notice the relationship between those two numbers, and that your probability is far too big (probability of getting 0 or greater when the mean is -10 is the warning sign)

You're looking for P(Z>z) and you have read off the P(Z=<z) value from tables...

Reply 12

Super199

Original post by SeanFM

I just chucked it into the calc. Do you have a casio fx-991e plus by any chance?

Reply 13

Kevin De Bruyne

Original post by Super199

I just chucked it into the calc. Do you have a casio fx-991e plus by any chance?

Afraid not :lol:

but your calculator most likely has found

P(Z \leq z)

.

Another hint is that 0.9408 + 0.0592 = 1.

Reply 14

Super199

Original post by SeanFM

Afraid not :lol:

but your calculator most likely has found

P(Z \leq z)

.

Another hint is that 0.9408 + 0.0592 = 1.

Lol no worries. Guess I will stick to the tables.

Reply 15

Kevin De Bruyne

Original post by Super199

Lol no worries. Guess I will stick to the tables.

You can use your calculator if you want, you just need to know what it's actually calculating.

Your calculation to find 0.9408 is correct - you just need 1 final calculation/step to get the correct answer.