Answer to Question #248429 in Statistics and Probability for Gillian

Question

For a normally distributed random variable, the standard deviation is 3.7. What is the mean if 10% of the distribution is less than 31?

Answer 1

Let m be unknown mean and

Y=(X-m)/3.7. Then Y=N(0,1)-standard normal

X=3.7$\cdot$ Y+m

P(X<31)=0.1=P(3.7$\cdot$ Y+m<31)=P(Y<(31-m)/3.7);

(31-m)/3.7=z(0,1)

For finding z score by p value p=0.1 we use software

(31-m)/3.7=-1.28

m=31+3.7$\cdot$ 1.28=35.736

Thus mean =35.736