The distribution of the number of viewers for the American I

The distribution of the number of viewers for the American Idol television show follows a normal distribution with a mean of 20 million with a standard deviation of 4 million.

Have between 25 and 28 million viewers? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

Have at least 17 million viewers? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

Exceed 32 million viewers? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

Solution

a.
P(25 < x < 28)
= P(x < 28) - P(x < 25)
= P(Z < (28 - 20) /4) - P(Z < (25 - 20) / 4)
= P(Z < 8 / 4) - P(Z < 5 / 4)
= phi(2) - phi(1.25)
= 0.9772 - 0.8925
= 0.0847

b.
P(x >= 17)
= 1 - P(x < 17)
= 1 - P(Z < (17 - 20) / 4)
= 1 - P(Z < -3/4)
= 1 - [ 1 - P(Z < 3/4) ]
= phi(3/4)
= 0.7734

c.
P(x > 32) = 1 - P(x <= 32) = 1 - P(Z <= (32 - 20) / 4) = 1 - phi(3) = 1 - 0.9987 = 0.0013