Question

a.) The J.O. Supplies Company buys calculators from a Korean supplier. The probability of a defective calculator is 30%. If 14 calculators are selected at random, what is the probability that more than 6 of the calculators will be defective?

b.) An important part of the customer service responsibilities of a cable company relates to the speed with which trouble in service can be repaired. Historically, the data show that the likelihood is 0.70 that troubles in a residential service can be repaired on the same day. For the first six troubles reported on a given day, what is the probability that: Fewer than 3 troubles will be repaired on the same day?

c.) Given the length an athlete throws a hammer is a normal random variable with mean 60 feet and standard deviation 2.5 feet, what is the probability he throws it between 55 feet and 65 feet?

d.) If x is a binomial random variable where n = 100 and p = 0.30, find the probability that x is more than 25 using the normal approximation to the binomial. Check the condition for continuity correction.

Answer 1

Part a)

X ~ B ( n = 14 , P = 0.3 )

P ( X >6 ) = 1 - P ( X <= 6 ) = 1 - 0.9067 = 0.0933

Part b)

X ~ B ( n = 6 , P = 0.7 )

P ( X < 3 ) = 0.0704

Part c)

X ~ N ( µ = 60 , σ = 2.5 )
P ( 55 < X < 65 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 55 - 60 ) / 2.5
Z = -2
Z = ( 65 - 60 ) / 2.5
Z = 2
P ( -2 < Z < 2 )
P ( 55 < X < 65 ) = P ( Z < 2 ) - P ( Z < -2 )
P ( 55 < X < 65 ) = 0.9772 - 0.0228
P ( 55 < X < 65 ) = 0.9545

Part d)

Mean = n * P = ( 100 * 0.3 ) = 30
Variance = n * P * Q = ( 100 * 0.3 * 0.7 ) = 21
Standard deviation = √(variance) = √(21) = 4.5826

P ( X > 25 )
Using continuity correction
P ( X > n + 0.5 ) = P ( X > 25 + 0.5 ) = P ( X > 25.5 )

X ~ N ( µ = 30 , σ = 4.5826 )
P ( X > 25.5 ) = 1 - P ( X < 25.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 25.5 - 30 ) / 4.5826
Z = -0.98
P ( ( X - µ ) / σ ) > ( 25.5 - 30 ) / 4.5826 )
P ( Z > -0.98 )
P ( X > 25.5 ) = 1 - P ( Z < -0.98 )
P ( X > 25.5 ) = 1 - 0.1635
P ( X > 25.5 ) = 0.8365