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:

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) =![[\frac{8-8.3}{1.4/ \sqrt{47}}< \frac{X'-u}{s.d/ \sqrt{n}} < \frac{9-8.3}{1.4/ \sqrt{47}}]](https://tex.z-dn.net/?f=%20%5B%5Cfrac%7B8-8.3%7D%7B1.4%2F%20%5Csqrt%7B47%7D%7D%3C%20%5Cfrac%7BX%27-u%7D%7Bs.d%2F%20%5Csqrt%7Bn%7D%7D%20%3C%20%5Cfrac%7B9-8.3%7D%7B1.4%2F%20%5Csqrt%7B47%7D%7D%5D)
= P(-1.47 <Z< 6.366)
= P( Z< 6.366) - P(Z< -1.47)
Using normal distribution table,

0.9999 - 0.0708
= 0.9291
c) Less than 7.5 minutes:
P(X'<7.5) = ![P [Z< \frac{7.5-8.3}{1.4/ \sqrt{47}}]](https://tex.z-dn.net/?f=%20P%20%5BZ%3C%20%5Cfrac%7B7.5-8.3%7D%7B1.4%2F%20%5Csqrt%7B47%7D%7D%5D%20)
P(X' < 7.5) = P(Z< -3.92)
NORMSDIST (-3.92)
= 0.0000
Answer:
C
Step-by-step explanation:
Becuz 9+9=17, which is the sum of 9 and 9.
C. Associative property of communication.
Answer:
We can conclude that the setting is stationed somewhere in the north where it's very cold, so cold that they can use sled dogs. We can also assume it's winter time because it normally snows during the winter.
So the setting is in the north where it is cold, during the winter time.