**Question 6: Grocery Spending in a One-Person Household****Problem Statement:**If a one-person household spends an average of $40 per week on groceries, find the maximum and minimum dollar amount spent per week for the middle 50% of one-person households. Assume that the standard deviation is $5 and the variable is normally distributed.**Explanation of Terms:**1. **Average**: The mean or central value of the dataset.2. **Standard Deviation**: A measure of the amount of variation or dispersion in a set of values.3. **Normal Distribution**: A probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence.**Solution Approach:**1. Given data: - Mean ($\mu$) = $40 - Standard deviation ($\sigma$) = $52. **Find the Z-scores for the middle 50%**: - The middle 50% means we are interested in the 25th percentile ($P_{25}$) and the 75th percentile ($P_{75}$). - Looking up the Z-scores for these percentiles: - $P_{25}$: Z=-0.6745 - $P_{75}$: Z=0.67453. **Calculate the minimum and maximum spending amounts**: - Minimum spending amount = $\mu + (Z_{25} \times \sigma)$ - Maximum spending amount = $\mu + (Z_{75} \times \sigma)$ Minimum: $40 + (-0.6745 \times 5) = 40 - 3.3725 = \$36.63$ Maximum: $40 + (0.6745 \times 5) = 40 + 3.3725 = \$43.37$**Answer:** For the middle 50% of one-person households, the minimum amount spent per week on groceries is approximately $36.63, and the maximum amount is approximately $43.37.

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Answered: 6. If a one-person household spends an…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

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6. If a one-person household spends an average of $40 per week on groceries, find the maximum and minimum dollar amount spent per week for the middle 50% of one-person households. Assume that the standard deviation is $5 and the variable is normally distributed.