A kids pool had a small opening
2 answers:
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Answer:
a) 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.
b) 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.
c) 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
.
Normal probability distribution
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:
The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
.
a. Find the probability that a randomly selected woman has a foot length less than 10.0 in
This probability is the pvalue of Z when .
has a pvalue of 0.7881.
So there is a 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.
b. Find the probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.
This is the pvalue of Z when X = 10 subtracted by the pvalue of Z when X = 8.
When X = 10, Z has a pvalue of 0.7881.
For X = 8:
has a pvalue of 0.0007.
So there is a 0.7881 - 0.0007 = 0.7874 = 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.
c. Find the probability that 25 women have foot lengths with a mean greater than 9.8 in.
Now we have .
This probability is 1 subtracted by the pvalue of Z when . So:
has a pvalue of 0.9772.
There is a 1-0.9772 = 0.0228 = 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.