91Ó°ÊÓ

The formula used $z=\frac{x−mean}{s\tan darddeviation}$

Only $5(13,13,13,15,19)$of the $20$pupils in the female distribution have pairs of shoes that are less than $22$. As a result, the percentile is calculated as follows: $5/20=0.25=25\%$As a result, the girl with $22$pairs of shoes is in the $25\%$ile. To put it another way, $25\%$of female students have fewer than$22$pairs of shoes. As a result,$25\%$ of the girl students have less than 22 pairs of shoes.

From the above-mentioned sorted list, $17$ of the $20$ male students had pairs of shoes that are below $2$. As a result, the appropriate percentile is calculated as follows: $17/20=0.85=85\%$

As a result, the youngster in the 85th percentile has $22$ pairs of shoes. To put it another way, $85\%$ of male students have fewer than $22$ pairs of shoes. As a result, $85\%$ of the male students have less than $22$ pairs of shoes.

The male is more unique because only $15\%$ of boys have more than $22$ pairs of shoes, but the girl's value is more evenly distributed over the distribution. Only $25\%$ of girls have fewer, whereas $75\%$ have the same number or more. Therefore, the boy is more unusual.