Answer:
Q1. 13 passengers
Q2. 0.1756 (approx. 0.18)
Step-by-step explanation:
Q1. 267 seats are available on the plane
5% is expected to fail to show up
Hence, no of passengers expected not to show up = 267 * 0.05
= 13.35 (approx 13 passengers)
Q2. See working in the attachment as I had to explain it step by step.
Answer:
a_n = 2.2 + 0.6 n
a_50 = 32.2
Step-by-step explanation:
What's the common difference of this series?
Common difference = .
Expression for the nth term:
n = 50 for the fiftieth term. Therefore
.
Answer:
Each adult male has a 5.05% probability of having a hip width greater than 16.2 inches.
There is a 0.01% probability that the 110 adult men will have an average hip width greater than 16.2 inches.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by
After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.
In this problem
Assume that adult men have hip breadths that are normally distributed with a mean of 14.4 inches and a standard deviation of 1.1 inches. This means that .
What is the probability that any one of those adult male will have a hip width greater than 16.2 inches?
For each one of these adult males, the probability that they have a hip width greater than 16.2 inches is 1 subtracted by the pvalue of Z when X = 16.2. So:
has a pvalue of 0.9495.
This means that each male has a 1-0.9495 = 0.0505 = 5.05% probability of having a hip width greater than 16.2 inches.
For the average of the sample
What is the probability that the 110 adult men will have an average hip width greater than 16.2 inches?
Now, we need to find the standard deviation of the sample before using the zscore formula. That is:
.
Now
has a pvalue of 0.9999.
This means that there is a 1-0.9999 = 0.0001 = 0.01% probability that the 110 adult men will have an average hip width greater than 16.2 inches.
