Please help. I’ve already spent 10mins trying to answer this.

Question

X is a normally distributed random variable with mean 61 and standard deviation 23. What is the probability that X is between 43 and 79?

(A) 56.61%

(B) 50.65%

(C) 46.41%

(D) 50.41%

Answer 1

How one answers this depends upon what one is using to provide normal distribution values.

In Excel, you would use =NORM.DIST(79,61,23,1)-NORM.DIST(43,61,23,1) = .5661

The old way was to use the normal table in the book and have P(z <= (79-61)/23) - P(z <= (43-61)/23) =

P(z <= .78) - P(z <= -.78) = .7823 - .2177 = .5646, but given that you have the exact value, the table method can't be what was expected.