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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.
ScoresFrequency
86 - 9614
97 - 1077
108 - 11813
119 - 12910
130 - 14011
141 - 1515

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.


Correct answer : (2)
 2.  
The net worth (in millions of dollars) of 30 business units in a city are shown in the frequency distribution.
Net worthFrequency
15 - 253
26 - 3610
37 - 4712
48 - 585
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.]


Correct answer : (3)
 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 loadNo. of cables
9.3 - 9.73
9.8 - 10.25
10.3 - 10.711
10.8 - 11.216
11.3 - 11.714
11.8 - 12.27
12.3 - 12.71
12.8 - 13.23

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.]


Correct answer : (1)
 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.


Correct answer : (1)
 5.  
The mean of the frequency distribution shown is 18.5. Find the value of t.
x1823t1516
f29487

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.


Correct answer : (1)
 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.


Correct answer : (2)
 7.  
Sixty eight randomly selected refrigerators were tested to determine their life span (in years). The relevant frequency distribution is as shown.
Lifetime of refrigeratorsFrequency
10.5 - 14.56
14.5 - 18.512
18.5 - 22.525
22.5 - 26.517
26.5 - 30.58

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.


Correct answer : (1)
 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 groupNumber of people
10 - 159
15 - 2022
20 - 2535
25 - 3028
30 - 3531
35 - 4017

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.


Correct answer : (1)
 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.
AgePercentage of house owners
15-2417.9
25-3425.6
35-4416.2
45-5430.3
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.


Correct answer : (3)
 10.  
The net worth (in million dollars) of 60 export oriented quality leather manufacturing units in a large state are distributed as shown.
Net worthLeather units
10 - 205
21 - 3110
32 - 4213
43 - 537
54 - 6418
65 - 757

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.


Correct answer : (4)

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