Answer:
74.22% probability that a family of four will spend more than $400.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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:
[tex]\mu = 442.54, sigma = 65[/tex]
What is the probability that a family of four will spend more than $400?
This is 1 subtracted by the pvalue of Z when X = 400. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{400 - 442.54}{65}[/tex]
[tex]Z = -0.65[/tex]
[tex]Z = -0.65[/tex] has a pvalue of 0.2578.
So there is a 1-0.2578 = 0.7422 = 74.22% probability that a family of four will spend more than $400.