Question

The U.S. Energy Information Administration reports that 36% of households in the U.S. use automatic dishwashers. What is the probability that from 40 randomly selected households that between 18 and 23 of them inclusively, use an automatic dishwasher? Give your answer to four decimal places.​

Answer 1

P :- Pobability of U.S  households use automatic dishwashers. = 0.36

Q :- Pobability of U.S  households does not use automatic dishwashers. = 1 - 0.36 = 0.64

Sample size n = 40

Using Normal Approximation

Checking the condition for n*P >=10 and n*Q >= 10

n * P = 40 * 0.36 = 14.4

n = Q = 40 * 0.64 = 25.6

Using continuity correction P ( n - 0.5 < X < n + 0.5 )

P ( 18 < X < 23 ) = P ( 18 - 0.5 < X < 23 + 0.5 )

P ( 17.5 < X < 23.5 )

Where, Mean = n * P = 40 * 0.36 = 14.4

Variance (X) = n * P * Q = 40 * 0.36 * 0.64 = 9.216

Standard deviation (X) =

P ( 17.5 < X < 23.5 )

Standardizing the value

Z = ( 17.5 - 14.4 ) / 3.0358

Z = 1.02

Z = ( 23.5 - 14.4 ) / 3.0358

Z = 3

P ( 1.02 < Z < 3 )

P ( 17.5 < X < 23.5 ) = P ( Z < 3 ) - P ( Z < 1.02 )

P ( 17.5 < X < 23.5 ) = 0.9986 - 0.8464

P ( 17.5 < X < 23.5 ) = 0.1522