Question

Given a random variable X having a normal distribution with μ = 60 and σ = 9, find the value of x that has 74% of the normal curve area to the left. Round your z-score and your final answer to 2 decimals.

Answer 1

P (Z<1.56)= 0.94

Explanation:

Step 1: Sketch the curve.

The probability that X<74 is equal to the blue area under the curve.

Step 2:

Since μ=60 and σ=9 we have:

P ( X<74 )=P ( X−μ<74−60 )=P (X−μ/σ<74−60/9)

Since x−μ/σ=Z and 74−60/9=1.56 we have:

P (X<74)=P (Z<1.56)

Step 3: Using the standard normal table to conclude that:

P (Z<1.56)=0.9406 , rounding off to 2 decimals we get P (Z<1.56)= 0.94