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# Statistics Worksheets - Page 2

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11.
Find the percentile rank for a weight of 92 pounds if the weights (in pounds) of 6 students are given:
97, 78, 86, 82, 88, 92
What value corresponds to the 60th percentile?
 a. 25th percentile, 86 b. 75th percentile, 88 c. 75th percentile, 86 d. 75th percentile, 90

#### Solution:

Write the weights of students in increasing order.
78, 82, 86, 88, 92, 97

Percentile rank for a weight of 92, p = (number of students below 92)+0.5Total number of students × 100%

Since, there are 4 values below a weight of 92 pounds, the percentile is 4+0.5 / 6 × 100% = 75th percentile.

So, the student whose weight is 92 pounds has a weight more than the weight of 75% of the students.

Position of the weight in order corresponds to the 60th percentile, c = n.p100, where n is the number of students, p the percentile.

c = 6×60 / 100 = 3.6 4, rounded to the next whole number.

So, the weight corresponding to the 60th percentile is 88 pounds.
[Start at the lowest value and count over to the 4th value.]

12.
The distance (in miles) run by 8 athletes per day are given. Find the value that corresponds to the 40th percentile.
4.2, 5.8, 4.5, 5.1, 3.8, 4.6, 3.2, 5.4, 4.3, 4.4
 a. 4.2 miles b. 3.8 miles c. 4.55 miles d. 4.35 miles

#### Solution:

Write the distances run by athletes in increasing order.
3.2, 3.8, 4.2, 4.3, 4.4, 4.5, 4.6, 5.1, 5.4, 5.8

To determine the distance corresponding to the 40th percentile, compute c = n.p100, where n is the number of athletes, p the percentile.

c = 10×40 / 100 = 4, which is a whole number.

So, the distance in miles corresponding to the 40th percentile is mean of 4th and 5th distances in the order; which is 4.3+4.4 / 2 = 4.35.

So, 4.35 miles corresponds to the 40th percentile.

13.
The height (in feet) of 8 waterfalls in different parts of the world are as follows:
1904, 1215, 2307, 2000, 1612, 2800, 2425, 1170.
Find the interquartile range.
 a. 1413.5 feet b. 2366 feet c. 3779.5 feet d. 952.5 feet

#### Solution:

Arrange the heights of waterfalls in increasing order.
1170, 1215, 1612, 1904, 2000, 2307, 2425, 2800

Median = 1952.
[Median is the mean of the middle values of the data set in the order.]

So, Q2 = 1952
[Q2 is the median.]

The median of the data values less than 1952 (median) is Q1.

The heights that are less than 1952 (median) are 1170, 1215, 1612, 1904.

Q1 = 1215+1612 / 2= 2827 / 2= 1413.5

The median of the data values greater than 1952 (median) is Q3.

The heights that are greater than 1952 (median) are 2000, 2307, 2425, 2800.

Q3 = 2307+2425 / 2= 4732 / 2= 2366

Interquartile range (IQR) = Q3 - Q1 = 2366 - 1413.5 = 952.5 feet.

14.
The life times (in weeks) of seven randomly selected bulbs are as recorded. Find midquartile range.
52, 63, 107, 96, 118, 74, 86
 a. 85 weeks b. 22 weeks c. 170 weeks d. 86 weeks

#### Solution:

Arrange the life times of bulbs in increasing order.
52, 63, 74, 86, 96, 107, 118

Median = 86.
[Median is the middle value of the data set in the order.]

So, Q2 = 86
[Q2 is the median.]

The median of the data values less than 86 (median) is Q1.

The life times that are less than 86 (median) are 52, 63, 74.

Q1 = 63.

The median of the data values greater than 86 (median) is Q3.

The life times that are greater than 86 (median) are 96, 107, 118.

Q3 = 107.

Midquartile range = Q1+Q32 = 63+107 / 2 = 170 / 2 = 85 weeks.

15.
The number of credits in business courses that 5 job applicants took is shown as 33, 27, 46, 54, 94. Check for outlier.
 a. 27 b. 63.5 c. no outlier d. 94

#### Solution:

Arrange the number of credits in increasing order.
27, 33, 46, 54, 94

Q2 = 46
[Q2 is the median, which is the middle value of the data set in the order.]

Q1 = 27+33 / 2 = 30
[Q1 is the median of the data values less than Q2.]

Q3 = 54+94 / 2 = 74
[Q3 is the median of the data values greater than Q2.]

Interquartile Range (IQR) = Q3 - Q1 = 74 - 30 = 44

To find the outlier, compute the cut-off points for outliers.

Lower fence = Q1 - 1.5(IQR) = 30 - 1.5(44) = - 36

Upper fence = Q3 + 1.5(IQR) = 74 + 1.5(44) = 140

If a data value is less than the lower fence or greater than the upper fence, then it is considered an outlier.

There are no values that are less than - 36 or greater than 140.

So, there is no outlier.

16.
The number of days taken by a company to manufacture different sizes of same product are given as 5, 6, 22, 15, 13, 18, 12, 50. Check for outliers.
 a. 5 b. 50 c. 5 & 6 d. 22 and 50

#### Solution:

Arrange the number of days taken in increasing order.
5, 6, 12, 13, 15, 18, 22, 50

Q2 = 13+15 / 2 = 14
[Q2 is the median, which is the mean of the middle values of the data set in the order.]

Q1 = 6+12 / 2 = 9
[Q1 is the median of the data values less than Q2.]

Q3 = 18+22 / 2 = 20
[Q3 is the median of the data values greater than Q2.]

Interquartile Range (IQR) = Q3 - Q1 = 20 - 9 = 11

To find the outlier, compute the cut-off points for outliers.

Lower fence = Q1 - 1.5(IQR) = 9 - 1.5(11) = - 7.5

Upper fence = Q3 + 1.5(IQR) = 20 + 1.5(11) = 36.5

If a data value is less than the lower fence or greater than the upper fence, then it is considered an outlier.

The value 50 is greater than 36.5.

So, 50 is the outlier.

17.
The number of inches of rainfall received in a year in 7 selected cities is given as 5.5, 5.8, 6.2, 7.9, 5.3, 6.8, 15.5. Check for outlier.
 a. 15.5 b. no outlier c. 5.3 d. 7.9 and 15.5

#### Solution:

Arrange the number of inches of rainfall received in increasing order.
5.3, 5.5, 5.8, 6.2, 6.8, 7.9, 15.5.

Q2 = 6.2
[Q2 is the median, which is the middle value of the data set in the order.]

Q1 = 5.5
[Q1 is the median of the data values less than Q2.]

Q3 = 7.9
[Q3 is the median of the data values greater than Q2.]

Interquartile Range (IQR) = Q3 - Q1 = 7.9 - 5.5 = 2.4

To find the outlier, compute the cut-off points for outliers.

Lower fence = Q1 - 1.5(IQR) = 5.5 - 1.5(2.4) = 1.9

Upper fence = Q3 + 1.5(IQR) = 7.9 + 1.5(2.4) = 11.5

If a data value is less than the lower fence or greater than the upper fence, then it is considered an outlier.

The value 15.5 is greater than 11.5.

So, 15.5 is the outlier.

18.
Write the $z$ - score and the test that has a better relative position of the two tests?
 Test 1 X = 53 $\stackrel{‾}{X}$ = 61 $s$ = 8 Test 2 X = 112 $\stackrel{‾}{X}$ = 122 $s$ = 9
 a. - 1.1, Test 2 b. - 1.1, Test 1 c. - 1, Test 1 d. - 1, Test 2

#### Solution:

For Test 1, z = X-Xs = 53-61 / 8 = - 1

For Test 2, z = X-Xs = 112-122 / 9 = - 1.1 < - 1

The score for Test 1 is relatively better then the score for Test 2.

19.
Find the $z$ scores of three students who had a score of 93, 80, 84 in an examination for a pre-medicine course which has a mean of 84 and a standard deviation of 4.
 a. 9, - 4, 0 b. 23.25, 20, 21 c. 2.25, - 1, 0 d. - 2.25, 1, 0

#### Solution:

z = score - meanstandard deviation = X-Xs = 93-84 / 4 = 2.25

z = X-Xs = 80-84 / 4 = - 1

z = X-Xs = 84-84 / 4 = 0

20.
The number of clients visiting a stock market over a 11-day period are as follows:
33, 38, 43, 30, 29, 40, 51, 27, 42, 23, 31
Find the interquartile range.
 a. 37 b. 4 c. 13 d. 28

#### Solution:

Arrange the number of clients visiting the stock market in increasing order.
23, 27, 29, 30, 31, 33, 38, 40, 42, 43, 51

Median = 33.
[Median is the middle value of the data set in the order.]

So, Q2 = 33
[Q2 is the median.]

The median of the data values less than 33 (median) is Q1.

The number of clients visiting the stock market that are less than 33 (median) are 23, 27, 29, 30, 31.

Q1 = 29

The median of the data values greater than 33 (median) is Q3.

The number of clients visiting the stock market that are greater than 33 (median) are 38, 40, 42, 43, 51.

Q3 = 42

Interquartile range (IQR) = Q3 - Q1 = 42 - 29 = 13.