Westonci.ca is your trusted source for finding answers to a wide range of questions, backed by a knowledgeable community. Experience the ease of finding accurate answers to your questions from a knowledgeable community of professionals. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
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:
[tex]mean=\frac{(4*6)+(11*5)+(11*6)+(9*7)+(8*8)+(9*9)+(10*7)}{61}[/tex]Solving further:
[tex]\begin{gathered} mean=\frac{423}{61} \\ \\ mean=6.9 \end{gathered}[/tex]The mean is 6.9
• (b). The median
The median is the middle value.
To find the median, we have:
[tex]\frac{n+1}{2}=\frac{61+1}{2}=\frac{62}{2}=31[/tex]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:
[tex]s=\sqrt{\frac{\Sigma f(x-x^{\prime})^2}{n-1}}[/tex]Thus, we have:
[tex]\begin{gathered} s=\sqrt{\frac{6(4-6.9)^2+11(5-6.9)^2+11(6-6.9)^2+9(7-6.9)^2+8(8-6.9)^2+9(9-6.9)^2+7(10-6.9)^2}{61-1}} \\ \\ s=\sqrt{3.59683} \\ \\ s=1.9 \end{gathered}[/tex]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:
[tex]\frac{30}{2}=15[/tex]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:
[tex]32+15=47[/tex]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:
[tex]\begin{gathered} \frac{61-5}{61}*100 \\ \\ =\frac{56}{61}*100 \\ \\ =0.9180*100 \\ \\ =91.8\text{ \%} \end{gathered}[/tex]• (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.
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.