Question

According to an NRF survey conducted by BIGresearch, the average family spends about $237 on electronics (computers, cell phones, etc.) in back-to-college spending per student. Suppose back-to-college family spending on electronics is normally distributed with a standard deviation of $54. If a family of a returning college student is randomly selected, what is the probability that: (a) They spend less than $160 on back-to-college electronics

Answer 1

Answer:

7.64% probability that they spend less than $160 on back-to-college electronics

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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:

$\mu = 237, \sigma = 54$

Probability that they spend less than $160 on back-to-college electronics

This is the pvalue of Z when X = 160. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{160 - 237}{54}$

$Z = -1.43$

$Z = -1.43$ has a pvalue of 0.0763

7.64% probability that they spend less than $160 on back-to-college electronics