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- Jun 18, 2026
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A grant writer for a local school district is trying to justify an application for funding an after-school program for low-income families. Census information for the school district shows an average household income of $26,200 with a standard deviation of $8,960. Assuming that the household income is normally distributed, what is the percentage of households in the school district with incomes of less than $12,000?
Explanation
Z = ($12,000 – $26,200) / $8,960 = –1.58.
From the table of areas under the standard normal curve, 5.71% of observations are more than 1.58 standard deviations below the mean.
My question is why do you not take 5.71% and divide by 2 since you are looking for the % below $12,000? Doesn’t the 5.71% mean that percent is the combined total on either side?
Explanation
Z = ($12,000 – $26,200) / $8,960 = –1.58.
From the table of areas under the standard normal curve, 5.71% of observations are more than 1.58 standard deviations below the mean.
My question is why do you not take 5.71% and divide by 2 since you are looking for the % below $12,000? Doesn’t the 5.71% mean that percent is the combined total on either side?