# Standard Normal Distribution Worksheet

Standard Normal Distribution Worksheet
• Page 1
1.
The scores of 60 students of a class are shown in the frequency distribution.
 Scores Frequency 86 - 96 14 97 - 107 7 108 - 118 13 119 - 129 10 130 - 140 11 141 - 151 5

Find the mean and the modal class of the data.
 a. 115 and 151 b. 115.2 and 86 - 96 c. 115.3 and 141 - 151 d. 115.2 and 14

#### Solution:

The midpoint of each class (scores) and the product with the frequency is as shown in the table.

Mean (X) = 1n f X = 6912 / 60 = 115.2
[Use n = f.]

The modal class is 86 - 96, since it has the largest frequency.

The mean of the scores of 60 students is 115.2 and the modal class of the data is 86 - 96.

2.
The net worth (in millions of dollars) of 30 business units in a city are shown in the frequency distribution.
 Net worth Frequency 15 - 25 3 26 - 36 10 37 - 47 12 48 - 58 5
Find the mean and the mode of the data and compare them.
 a. 38, 37, mean > mode b. 284.75, 284.75, mean = mode c. 37.97, 42, mean < mode d. 284.75, 37, mean > mode

#### Solution:

The midpoint of each class (net worth) and the product with the frequency is as shown in the table.

Mean (X) = 1n f X = 1139 / 30 = 37.97
[Use n = f.]

The modal class is 37 - 47, since it has the largest frequency. The mode of the date is 37+472 = 42.

The mean of the net worth (in millions of dollars) of 30 business units is 37.97 and the mode of the data is 42.

So, mean < mode.
[Compare the values of mean and mode.]

3.
The distribution of maximum load in short tons (1 short ton = 2000 lb) supported by certain cables produced by a company is shown. Find the mode and mean of the maximum load of the cables. Which is greater, mean or mode?
 Maximum load No. of cables 9.3 - 9.7 3 9.8 - 10.2 5 10.3 - 10.7 11 10.8 - 11.2 16 11.3 - 11.7 14 11.8 - 12.2 7 12.3 - 12.7 1 12.8 - 13.2 3

 a. 11, 11.108, mean b. 10.8, 11.108, mean c. 11.108, 11, mode d. 11.2, 11.108, mode

#### Solution:

The modal class is 10.8 - 11.2, since it has the largest frequency. The mode of the data is 10.8+11.22 = 11.

The midpoint of each class (maximum load) and the product with the frequency is as shown in the table.

Mean (X) = 1n f X = 666.5 / 60 = 11.108
[Use n = f.]

The mean is 11.108 and the mode of the data is 11.

So, mean > mode.
[Compare the values of mean and mode.]

4.
The average score of students in statistics examination is 90, with a variance of 4. The average score of students in algebra examination is 102, with a variance of 7. Which course is more variable in terms of scores?
 a. algebra b. both are equally variable c. cannot be determined d. statistics

#### Solution:

Coefficient of variation of statistics class = 490 · 100% = 2.22%
[C.V = Standard deviationMean · 100%.]

Coefficient of variation of algebra class = 7102 · 100% = 2.59%

Since, the coefficient of variation is larger for algebra class, the scores of students in algebra course is more variable than the scores of students in statistics course.

5.
The mean of the frequency distribution shown is 18.5. Find the value of $t$.
 $x$ 18 23 $t$ 15 16 $f$ 2 9 4 8 7

 a. 20 b. 30.6 c. 20.6 d. 30

#### Solution:

Mean of the distribution is 18.5.

1n X f = 18.5

18.5 = 130(475 + 4t)

555 = 475 + 4t

80 = 4t

20 = t

The value of t is 20.

6.
The average age of technicians at B & T company is 26 years, with a standard deviation of 6 years. The average salary of the technicians is $31,000 with a standard deviation of$4000. Is the variation in the salary of technicians more than the variation in their age ?
 a. yes b. no c. cannot be determined d. both are equal

#### Solution:

Coefficient of variation of the age of technicians = 6 / 26 · 100%= 23.08%
[C.V = Standard deviationMean · 100%.]

Coefficient of variation of the salary of technicians = 4000 / 31000 · 100% = 12.9%

Since the coefficient of the variation is larger for the age of the technicians, the variation in the age is more than the variation in salary.

7.
Sixty eight randomly selected refrigerators were tested to determine their life span (in years). The relevant frequency distribution is as shown.
 Lifetime of refrigerators Frequency 10.5 - 14.5 6 14.5 - 18.5 12 18.5 - 22.5 25 22.5 - 26.5 17 26.5 - 30.5 8

Find the variance and standard deviation.
 a. 20.01, 4.47 b. 3.76, 14.2 c. 208.75, 14.45 d. 14.2, 201.64

#### Solution:

The midpoint of each class (lifetime of refrigerators) and the product with the frequency is as shown in the table.

Variance of the sample,
s2 = Σ fX2-[(Σ f X)2n]n-1
[Use n = f.]

= 31413-[(1430)268]67

= 31413-30072.0567

= 1340.95 / 67 = 20.01

So, the sample variance is 20.01.

Sample standard deviation, s = sample variance = 20.01 = 4.47

So, the sample standard deviation is 4.47.

The lifetime of sixty eight refrigerators has a sample variance of 20.01 and a sample standard deviation of 4.47.

8.
The number of people of different age groups visiting a recreation club in a town are as shown in the frequency distribution table.
 Age group Number of people 10 - 15 9 15 - 20 22 20 - 25 35 25 - 30 28 30 - 35 31 35 - 40 17

Find the mean of the data.
 a. 26.06 b. 142 c. 616.67 d. 3700

#### Solution:

The midpoint of each class (age group) and the product with the frequency is as shown in the table.

Mean = 1n f X = 3700 / 142 = 26.06
[Use n = f.]

The number of people of different age groups visiting a recreation club in a town has a mean of 26.06.

9.
The table represents the percentage of house owners and corresponding ages (in years) of a municipal area in the year 2000. Find the variance and standard deviation.
 Age Percentage of house owners 15-24 17.9 25-34 25.6 35-44 16.2 45-54 30.3
 a. 11.45 and 131.18 b. 129.73 and 11.39 c. 131.18 and 11.45 d. 11.45 and 3.38

#### Solution:

The midpoint of each class (Age) and the product with the frequency is as shown in the table.

Variance of the sample,
s2 = Σ fX2-[(Σ f X)2n]n-1
[Use n = f.]

= 128603.5-[(3244)290]89

= 128603.5-116928.1789

= 11675.32 / 89 = 131.18

So, the sample variance is 131.18.

Sample standard deviation, s = sample variance = 131.18 = 11.45

So, the sample standard deviation is 11.45.

The house ownership rates by the age of the householder in the year 2000 has a variance of 131.18 and a standard deviation of 11.45.

10.
The net worth (in million dollars) of 60 export oriented quality leather manufacturing units in a large state are distributed as shown.
 Net worth Leather units 10 - 20 5 21 - 31 10 32 - 42 13 43 - 53 7 54 - 64 18 65 - 75 7

Find the variance and standard deviation.
 a. 16.93 and 286.57 b. 45.07 and 6.71 c. 16.93 and 4.11 d. 286.57 and 16.93

#### Solution:

The midpoint of each class (net worth) and the product with the frequency is as shown in the table.

Variance of the sample,
s2 = Σ fX2-[(Σ f X)2n]n-1
[Use n = f.]

= 138768-[(2704)260]59

= 138768-121860.2659

= 16907.74 / 59 = 286.57

So, the sample variance is 286.57.

Sample standard deviation, s = sample variance = 286.57 = 16.93

So, the sample standard deviation is 16.93.

The net worth of 60 export oriented quality leather manufacturing units has variance of 286.57 and a standard deviation of 16.93.