Q:

Z-score and PercentileHelp please. Answer with explanation. Both circled numbers

Accepted Solution

A:
I assume all these measures are normally distributed, in which case the median and mean are equal.15a. [tex]P(X>66,650)=0.5[/tex] because the normal distribution is symmetric about the mean, and so 50% of families have an income greater than the average.15b. [tex]P(X>178,500)=0.1[/tex] by definition of percentile. The 90th percentile is $178,500, so the top 10% of families earn more than this.15c. Because it's continuous, this distribution has the property [tex]P(66,650<X<178,500)=P(X<178,500)-P(X<66,650)[/tex]. The top 10% earn more than $178,500, so the remaining 90% earn less. 50% of families earn more than the average, so 50% of families earn less than the average. Then [tex]P(66,650<X<178,500)=0.9-0.5=0.4[/tex] so 40% of families earn between the median and 90th percentile.16a. [tex]P(X>570)=0.75[/tex] or 75% because the first quartile is $570, meaning 25% of renters pay less than this amount.16b. [tex]P(X<708)=0.5[/tex] or 50% by the same reasoning used in (15a).16c. [tex]P(570<X<708)=P(X<708)-P(X<570)=0.5-0.25=0.25[/tex] or 25% by the same reasoning as in (15c).