Miguel is a baseball pitcher. This season, he struck
1 answer:
You might be interested in
Using the normal distribution, there is a 0.7357 = 73.57% probability that the sample proportion of households spending more than $125 a week is less than 0.31.
<h3>Normal Probability Distribution</h3>The z-score of a measure X of a normally distributed variable with mean and standard deviation
is given by:
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
- By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean
and standard deviation
, as long as
and
.
The estimate and the sample size are:
p = 0.29, n = 207.
Hence the mean and the standard error are given as follows:
.
.
The probability that the sample proportion of households spending more than $125 a week is less than 0.31 is the <u>p-value of Z when X = 0.31</u>, hence:
By the Central Limit Theorem:
Z = 0.63
Z = 0.63 has a p-value of 0.7357.
0.7357 = 73.57% probability that the sample proportion of households spending more than $125 a week is less than 0.31.
More can be learned about the normal distribution at brainly.com/question/4079902
#SPJ1
FOLLOW ME FOR CLEARING YOUR NEXT DOUBT
