Question

Let the random variable X follow a normal distribution with µ = 18 and σ2 = 11. Find the probability that X is greater than 10 and less than 17.

Answer 1

X follows normal distribution with mean = 18 and = 11

Probability that X is greater than 10 and less than 17 = P(10<X<17) = P(X<17) - P(X<10)

P(X<17)

Z-score for 17 = (17-)/ = (17-18)/3.3166 = -0.30

From standard normal tables, P(Z<-0.30) = 0.3821

P(X<17) = P(Z<-0.30) = 0.3821

P(X<10)

Z-score for 10 = (10-)/ = (10-18)/3.3166 = -2.41

From standard normal tables, P(Z<-2.41) = 0.0080

P(X<10) = 0.0080

P(10<X<17) = P(X<17) - P(X<10) = 0.3821 - 0.0080=0.3741

Probability that X is greater than 10 and less than 17 = 0.3741