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Normal Distribution Calculations

Total questions: 25

Worksheet time: 54mins

Name
Class
Date
1.

XX  is normally distributed with mean 12 and standard deviation 8.

Which image corresponds to P(X≥15)P\left(X\ge15\right)  ?

a)
b)
c)
2.

XX  is a random variable with normal distribution of mean 8 and standard deviation 1. Select the image corresponding to the curve of this normal distribution.

a)
b)
c)
3.
According to the empirical rule, how much of the data falls within 3 standard deviations? 
a)
25%
b)
68%
c)
95%
d)
99.7%
4.

Given a mean of 75 and a standard deviation of 7, what percentage of scores are between 75 and 82?

a)

68%

b)

81.5%

c)

16%

d)

34%

5.

According to the empirical rule, what percentage of data falls within 1 standard deviation of the mean?

a)

25%

b)

68%

c)

95%

d)

99.7%

6.

35 is

a)

mean

b)

variance

c)

standard deviation

d)

1st quartile

7.

What is the standard deviation on this Normal Curve?

a)

6

b)

12

c)

34

d)

18

8.

What is the value of the mean on this Normal Curve?

a)

6

b)

34

c)

52

d)

70

9.

Use the following information and the Empirical Rule to estimate the answer.


The ages of golfers are normally distributed, with a mean of 38 and a standard deviation of 4.


Find the percentage of golfers that are between 30 and 46 years old.

a)

68%

b)

94%

c)

95%

d)

99.7%

10.
The mean number of accidents a week at a company is 6.4 with a standard deviation of 1.5.  What proportion of weeks would you expect to have less than 5 accidents?
a)
0.6915
b)
0.1762
c)
0.8238
d)
-0.93
11.
The mean number of accidents a week at a company is 6.4 with a standard deviation of 1.5.  What proportion of weeks would you expect to have more than 7 accidents?
a)
0.3446
b)
0.2856
c)
0.4
d)
0.6554
12.
The mean GPA of students in a course at UCDavis is 3.2 with a standard deviation of 0.3. What percent of students in the course have a GPA between  2.9 and 3.8? 
a)
81.5%
b)
47.5%
c)
68%
d)
95%
13.
The average waist size for teenage males is 29 inches with a standard deviation of 1.4 inch. 
If waist sizes are normally distributed, estimate the proportion of teenagers who will have waist   sizes greater than 31 inches?
a)
92.4%
b)
10.5%
c)
16%
d)
7.6%
14.

Students pass a test if they score 50% or more.


The marks of a large number of students were sampled and the mean and standard deviation were calculated as 42% and 8% respectively.


Assuming this data is normally distributed, what percentage of students pass the test? in percentage...

a)

5

b)

16

c)

24

d)

32

15.

95% of students at school weigh between 62 kg and 90 kg.

Assuming this data is normally distributed, what are the mean and standard deviation?

a)

Mean = 66 kg

S.D. = 7 kg

b)

Mean = 76 kg

S.D. = 7 kg

c)

Mean = 86 kg

S.D. = 7 kg

d)

Mean = 76 kg

S.D. = 14 kg

16.

Given a mean of 75 and a standard deviation of 7, what percentage of scores are between 75 and 82?

a)

68%

b)

81.5%

c)

16%

d)

34%

17.

If this is the graph of data normally distributed with a mean of 30 and a standard deviation of 5, what label (number) is written at the blue circle?

a)

40

b)

35

c)

45

d)

50

18.

The marks obtained by the students are normally distributed with mean 61 marks and variance 100 marks. Find the probability of the students obtain at least 85 marks.

a)

0.4207

b)

0.0082

c)

0.1112

d)

0.9918

19.
This data is normally distributed.  What percent of the data is in the shaded region?
a)
5%
b)
95%
c)
2.5%
d)
50%
20.

This data is normally distributed. What percent of the data is in the shaded regions?

a)

5%

b)

95%

c)

2.5%

d)

50%

21.

This data is normally distributed. What percent of the data is in the shaded region?

a)

68%

b)

32%

c)

16%

d)

50%

22.

This data is normally distributed. What percent of the data is in the shaded region?

a)

68%

b)

95%

c)

99.8%

d)

50%

23.

Any normal probability distribution can be converted to the standard normal probability distribution by the following formula

a)

z=X−μσz=\frac{X-\mu}{\sigma}

b)

z=μ−Xσz=\frac{\mu-X}{\sigma}

c)

z=X+μσz=\frac{X+\mu}{\sigma}

d)

z=Xμσz=\frac{X\mu}{\sigma}

24.
a)

2%2\%  

b)

31%31\%  

c)

48%48\%  

d)

95%95\%  

25.
a)

0.22570.2257  

b)

0.27430.2743  

c)

0.72570.7257  

d)

0.77570.7757