Question

6. The age of licensed Michigan deer hunters has a normal distribution with a mean of 42 years and a standard deviation of 15 years, So, X ~ Normal(μ = 42, σ = 15). What percentage of Michigan deer hunter is between 18 years old and 30 years old? Round to 1 decimal place for the percentage.

Answer 1

μ = 42, σ = 15

We want to find P(18 < X < 30)

P(18 < X < 30) = P(x < 30) - P(x < 18)

first find P(x < 30)

z= −0.8

P(x < 30) = P(z < −0.8)

find P(z < −0.8) using normal z table we get

P(z < −0.8) = 0.2119

P(x < 30) = 0.2119

now find P(x < 18)

z= −1.6

P(x < 18) = P(z < −1.6)

find P(z < −1.6) using normal z table we get

P(z < −1.6) = 0.0548

P(x < 18) = 0.0548

P(18 < X < 30) = P(x < 30) - P(x < 18)

P(18 < X < 30) =0.2119−0.0548

P(18 < X < 30) =0.1571

Percentage = 15.7