b To make a histogram, draw a bar that will correspond to each value from the set. The height of a bar should be proportional to the number of times the corresponding value occurs in the set.
C
c The highest point of the normal curve should be at the mean of the values.
A
a Set 2
B
b
C
c
Practice makes perfect
a We are asked to choose from the three given sets of values one that appears to be distributed normally. Let's look at the given sets.
Set 1
Set 2
Set 3
1
5
5
10
7
6
5
7
9
19
7
1
2
4
1
7
11
5
1
7
11
7
7
1
2
7
10
10
9
4
6
7
2
9
7
8
To make things easier, let's sort the values in each set.
Set 1
Set 2
Set 3
1
4
1
1
5
1
2
7
1
2
7
2
5
7
4
6
7
5
7
7
5
7
7
6
9
7
8
10
7
9
10
9
10
19
11
11
The values in sets 1 and 3 are evenly distributed, so they are not normally distributed. In set 2, most part of the values is in the middle. Therefore, only the set 2 appears to be normally distributed.
b In Part A we came to the conclusion that set 2 is normally distributed. We will make a histogram of the values from set 2. First, let's sort the values in this set.
Set 2
4
5
7
7
7
7
7
7
7
7
9
11
The values range from 4 to 11. Let's count the number of times that each value occurs in the set.
Set 2
Value
Number of Times it Occurs
4
1
5
1
6
0
7
8
8
0
9
1
10
0
11
1
Now, let's make a bar graph. Each bar will represent a value and the height of each bar should be proportional to the number of times the corresponding value occurred in set 2.
c In Part B we created a bar graph.
Let's sketch a normal curve over the bar graph. The highest point of the curve should be at the mean, which is about 7.
