Answer:
75.17% probability that the sample proportion of households spending more than $125 a week is less than 0.32
Step-by-step explanation:
Problems of normally distributed samples are 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.
For the sampling distribution of a proportion p with size n, we have that
In this problem, we have that:
So
What is the probability that the sample proportion of households spending more than $125 a week is less than 0.32
This is the pvalue of Z when X = 0.32. So
has a pvalue of 0.7517
75.17% probability that the sample proportion of households spending more than $125 a week is less than 0.32