# Standard deviation / mean problem

• Jan 11th 2009, 04:42 PM
Standard deviation / mean problem
In a group of students the mean score for the SAT Math is 510 with a standard deviation of 95. The mean score for the SAT verbal is 490 with a standard deviation of 105.

a) Determine the mean combined score for this group of students.

b) Determine the combined standard deviation for this group of students.

c) What is the probability that one of these students achieves a combined score of 1400 or better?

I dont think you can add means together, obviously not if their standard deviations are different. Just dont know what to do on this one. Any help?
• Jan 11th 2009, 07:40 PM
mr fantastic
Quote:

Originally Posted by Bradley55
In a group of students the mean score for the SAT Math is 510 with a standard deviation of 95. The mean score for the SAT verbal is 490 with a standard deviation of 105.

a) Determine the mean combined score for this group of students.

b) Determine the combined standard deviation for this group of students.

c) What is the probability that one of these students achieves a combined score of 1400 or better?

I dont think you can add means together, obviously not if their standard deviations are different. Just dont know what to do on this one. Any help?

Let X be the random variable SAT math score and Y the random variable SAT verbal score.

Define U = X + Y.

(a) E(U) = E(X) + E(Y).

(b) IF X and Y are independent (a big IF but otherwise the question can't be answered) then Var(U) = Var(X) + Var(Y). Therefore sd(U) = ....

(c) You're probably meant to assume that X and Y follow a normal distribution. In which case U does to and has the above parameters. Calculate Pr(U > 1400).
• Jan 11th 2009, 07:47 PM
A) E(U) = 490+510 = 1000? combined mean = 1000?

b) Problem nowhere states anything about a normal curve or independent variables. So what should i do?

im still lost on b and c
• Jan 11th 2009, 08:21 PM
mr fantastic
Quote:

Originally Posted by Bradley55
A) E(U) = 490+510 = 1000? combined mean = 1000?

b) Problem nowhere states anything about a normal curve or independent variables. So what should i do?

im still lost on b and c

You need to get clarification from your instructor (who will probably sat to make those assumptions). Assuming independence, the standard deviation is (as the formula I gave in my previous reply clearly says) $\sqrt{95^2 + 105^2}$.
• Jan 11th 2009, 08:35 PM
So now that I have the combined mean and combined standard deviation, can i just do another normal cdf for C?

normalcdf(1400,9999999,1000,141.59)
• Jan 11th 2009, 08:48 PM
mr fantastic
Quote:

Originally Posted by Bradley55
So now that I have the combined mean and combined standard deviation, can i just do another normal cdf for C?

normalcdf(1400,9999999,1000,141.59)

Yes. And actually sd = 141.60 if you're going to use 2 decimal place accuracy.
• Jan 11th 2009, 09:06 PM
.0023 or .23% that one of these students achieves a combined score of 1400 or better!

Woot. You should teach at my school.
• Jan 11th 2009, 09:21 PM
mr fantastic
Quote:

Originally Posted by Bradley55
.0023 or .23% that one of these students achieves a combined score of 1400 or better!

Woot. You should teach at my school.

I don't have the patience to teach. When I do teach it is to small children - they are still full of wonder, treat you with respect and awe, and don't worry about getting into medicine or law school when they leave school.
• Jan 11th 2009, 09:24 PM
mr fantastic
Quote:

Originally Posted by Bradley55
.0023 or .23% that one of these students achieves a combined score of 1400 or better!

Woot. You should teach at my school.

If you're rounding to 4 decimal places then it's 0.0024.