Question

61 randomly selected students were asked the number of pairs of shoes they have. Let X represent the numberof pairs of shoes. The results are as follows:7 8 9 10# of Pairs of Shoes 4 5Frequency611 11 989 76Round all your answers to 4 decimal places where possible.The mean is:The median is:The sample standard deviation is:The first quartile is:The third quartile is:What percent of the respondents have at least 5 pairs of Shoes?28% of all respondents have fewer than how many pairs of Shoes?

Answer 1

Given the table:

Number of pairs of shoes 4 5 6 7 8 9 10

Frequency 6 11 11 9 8 9 7

Let's solve for the following:

• (a). The mean

To find the mean, we have:

`mean=((4*6)+(11*5)+(11*6)+(9*7)+(8*8)+(9*9)+(10*7))/(61)`

Solving further:

`\begin{gathered} mean=(423)/(61) \\ \\ mean=6.9 \end{gathered}`

The mean is 6.9

• (b). The median

The median is the middle value.

To find the median, we have:

`(n+1)/(2)=(61+1)/(2)=(62)/(2)=31`

The median will be the 31st number.

From the table, the 31st number is = 7

Therefore, the median is 7

• (C). The sample standard deviation

To find the standard deviation, apply the formula:

`s=\sqrt{(\Sigma f(x-x^(\prime))^2)/(n-1)}`

Thus, we have:

The standard deviation is 1.9

• (d). The first quartile.

The first quartile is the median of the lower half of the data.

To lower half of data is from 1 to 30th data.

The median of the lower half will be:

This means the value in the 15th frequency is the lower quartile.

The value in the 15th frequency is 5.

Therefore, the first quartile is 5.

• (e). Upper quartile

This is the median of the upper half of the data.

The median is the frequency from 32 to 61 frequency.

The median will be:

`32+15=47`

The 47th data.

The value of the 47th data is 9

Therefore, the third quartile is 9.

• (f). What percent have at least 5 pairs of shoes?

Only 6 respondents have less than 5 pairs of shoes.

To find the percent with at least 5 pairs of shoes, we have:

`\begin{gathered} (61-5)/(61)*100 \\ \\ =(56)/(61)*100 \\ \\ =0.9180*100 \\ \\ =91.8\text{ \%} \end{gathered}`

• (g). 28% have fewer than how many pairs of shoes?

28% have fewer than 6 pairs of shoes.

ANSWER:

• Mean = 6.9

,

• Median = 7

,

• Standard deviation = 1.9

,

• First quartile = 5

,

• Third quartile = 9

,

• At least 5 pairs = 91.8%

,

• 28% have fewer than, ,6 pairs of shoes.