Question

(a) A shipment of 120 fasteners that contains 4 defective fasteners was sent to a manufacturing plant. The quality-control manager at the manufacturing plant randomly selects 5 fasters and inspects them. What is the probability that exactly 1 fastener is defective? (b) The Energy Information Administration (EIA) sampled 900 retail gasoline outlets and reported that, the national average gasoline price for regular-grade gasoline to be $4.113 per gallon with a standard deviation of $0.11 per gallon. Construct a 95% Confidence Interval for mean price of regular-grade gasoline price.

Answer 1

a) Probability that exactly 1 fastener is defective, P(X = 1) = 0.144

b) Confidence interval for mean price,
`CI = [4.1058, 4.1202]`

Explanation:

a) Total number of fasteners = 120

Number of defective fasteners = 4

Probability of selecting a defective fastener, p = 4/120

p = 0.033

Probability of selecting an undefective fastener, q = 1 - p

q = 1 - 0.033

q = 0.967

5 fasteners were randomly selected, n =5

Probability that exactly one fastener is defective:

`P(X =r) = (nCr) p^r q^(n-r)\\P(X =1) = (5C1) 0.033^1 0.967^(5-1)\\P(X =1) = 0.144`

b) Number of gasoline outlets sampled, n = 900

Average gasoline price,
`\bar{x} = 4.113`

Standard deviation,
`\sigma = 0.11`

Confidence Level, CL = 95% = 0.95

Significance level,
`\alpha = 1 - 0.95 = 0.05`

`\alpha/2 = 0.05/2 = 0.025`

From the standard normal table,
`z_(\alpha/2) = z_(0.025) = 1.96`

error margin can be calculated as follows:

`e_(margin) = z_(\alpha/2) * (\sigma)/(√(n) ) \\e_(margin) = 1.96 * (0.11)/(√(900) )\\e_(margin) = 0.0072`

The confidence interval will be given as:

`CI = \bar{x} \pm e_(margin) \\CI = 4.113 \pm 0.0072\\CI = [(4.113-0.0072), (4.113+0.0072)]\\CI = [4.1058, 4.1202]`