According to the Bureau of Labor Statistics, in 2019 U.S. households spent an average of $\$ 386.92$ per month on food. Assume this data is normally distributed with a standard deviation of $\$ 112$. Use this information to answer the following question.
What Z-score would correspond to a household that spent an average of $\$ 555$ on food per month in 2019 ? Make sure to round your answer to 2 decimal places; i.e. if your answer was 87.654321 then you would type in 87.65 .
Answer
Final Answer: The Z-score that would correspond to a household that spent an average of \( \$ 555 \) on food per month in 2019 is \(\boxed{1.50}\)
Step 1 ：According to the Bureau of Labor Statistics, in 2019 U.S. households spent an average of \( \$ 386.92 \) per month on food. Assume this data is normally distributed with a standard deviation of \( \$ 112 \).
Step 2 ：We want to find the Z-score for a household that spent an average of \( \$ 555 \) on food per month in 2019.
Step 3 ：The Z-score is a measure of how many standard deviations an element is from the mean. The formula for calculating the Z-score is: \( Z = \frac{X - \mu}{\sigma} \)
Step 4 ：In this formula: \( X \) is the value we are comparing to the mean (in this case, \( \$ 555 \)), \( \mu \) is the mean (in this case, \( \$ 386.92 \)), and \( \sigma \) is the standard deviation (in this case, \( \$ 112 \)).
Step 5 ：Plugging these values into the formula, we get: \( Z = \frac{555 - 386.92}{112} \)
Step 6 ：Calculating the above expression, we get: \( Z = 1.5007142857142857 \)
Step 7 ：Rounding this to 2 decimal places, we get: \( Z = 1.50 \)
Step 8 ：Final Answer: The Z-score that would correspond to a household that spent an average of \( \$ 555 \) on food per month in 2019 is \(\boxed{1.50}\)
