Respuesta :
Answer:
a) Â Probability that exactly 1 fastener is defective, P(X = 1) = 0.144
b) Confidence interval for mean price, [tex]CI = [4.1058, 4.1202][/tex]
Step-by-step 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:
[tex]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[/tex]
b) Number of gasoline outlets sampled, n = 900
Average gasoline price, [tex]\bar{x} = 4.113[/tex]
Standard deviation, [tex]\sigma = 0.11[/tex]
Confidence Level, CL = 95% = 0.95
Significance level, [tex]\alpha = 1 - 0.95 = 0.05[/tex]
[tex]\alpha/2 = 0.05/2 = 0.025[/tex]
From the standard normal table, [tex]z_{\alpha/2} = z_{0.025} = 1.96[/tex]
error margin can be calculated as follows:
[tex]e_{margin} = z_{\alpha/2} * \frac{\sigma}{\sqrt{n} } \\e_{margin} = 1.96 * \frac{0.11}{\sqrt{900} }\\e_{margin} = 0.0072[/tex]
The confidence interval will be given as:
[tex]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][/tex]
