Answer:
The number is 8
Step-by-step explanation:
Using the Normal distribution, it is found that 0.0359 = 3.59% of US women have a height greater than 69.5 inches.
In a <em>normal distribution</em> with mean
and standard deviation
, the z-score of a measure X is given by:
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
US women’s heights are normally distributed with mean 65 inches and standard deviation 2.5 inches, hence
.
The proportion of US women that have a height greater than 69.5 inches is <u>1 subtracted by the p-value of Z when X = 69.5</u>, hence:



has a p-value of 0.9641.
1 - 0.9641 = 0.0359
0.0359 = 3.59% of US women have a height greater than 69.5 inches.
You can learn more about the Normal distribution at brainly.com/question/24663213
Every value{0.01, 0.29, 85%} can represent the probability of an event occurring except option d that is 1.5.
Given to us,
a.) 
b.) 0.29
c.) 85%
d.) 
The probability help us to know about the probability of specific events occurring.
For a sure event, the probability is always 1,
while for an event that will never happen the probability is always 0.
Thus, probability(p),
.
Now looking at the options,
a.)
= 0.01
b.) 0.29
c.) 85% = 0.85
d.)
= 1.5
Now comparing each option with
.
Therefore, the only option which is not feasible is option d that is 1.5.
Hence, every value{0.01, 0.29, 85%} can represent the probability of an event occurring except option d that is 1.5.
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brainly.com/question/795909
Answer:
4,055
Step-by-step explanation: