Complete question:
He amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and standard deviation 1.4 minutes. Suppose that a random sample of n equals 47 customers is observed. Find the probability that the average time waiting in line for these customers is
a) less than 8 minutes
b) between 8 and 9 minutes
c) less than 7.5 minutes
Answer:
a) 0.0708
b) 0.9291
c) 0.0000
Step-by-step explanation:
Given:
n = 47
u = 8.3 mins
s.d = 1.4 mins
a) Less than 8 minutes:
[tex]P(X<8) = P \frac{X'-u}{s.d/ \sqrt{n}} < \frac{8-8.3}{1.4/ \sqrt{47}}][/tex]
P(X' < 8) = P(Z< - 1.47)
Using the normal distribution table:
NORMSDIST(-1.47)
= 0.0708
b) between 8 and 9 minutes:
P(8< X' <9) =[tex] [\frac{8-8.3}{1.4/ \sqrt{47}}< \frac{X'-u}{s.d/ \sqrt{n}} < \frac{9-8.3}{1.4/ \sqrt{47}}][/tex]
= P(-1.47 <Z< 6.366)
= P( Z< 6.366) - P(Z< -1.47)
Using normal distribution table,
[tex] NORMSDIST(6.366)-NORMSDIST(-1.47) [/tex]
0.9999 - 0.0708
= 0.9291
c) Less than 7.5 minutes:
P(X'<7.5) = [tex] P [Z< \frac{7.5-8.3}{1.4/ \sqrt{47}}] [/tex]
P(X' < 7.5) = P(Z< -3.92)
NORMSDIST (-3.92)
= 0.0000
