Assume the standard deviation

Question

If a one-person household spends an average of $56 per week on groceries, find the maximum and minimum amounts spent per week for the middle 50% of one-person households. Assume the standard deviation is $10 and the variable is normally distributed. Round your answers to the nearest hundredth.

Minimum: $

Maximum: $

Answer 1

The middle 50% is between the 25th and 75th percentiles, which are the minimum and maximum values

use formula: z = (x - mean)/SD, where x is amount spent and mean = 56 and SD = 10

find z corresponding with 75th percentile. it is 0.674

using formula: 0.674 = (x - 56)/10

x = 10 * 0.674 + 56 = $62.74 is the maximum amount

since we are looking at middle 50%, the minimum is the same distance below the mean (56) that the maximum is above it, so:

minimum = 56 - (62.74 - 56) = $49.26