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MEAN. LET’S CHECK WHAT YOU KNOW. Simplify 3+2+45+67+78+96 = 291 (45+67+56+43+23)= 78 3 3. 9.261 = 3.087 3 4. 255 ÷ 4 = 63.75 5. 137 ÷ 5 = 27.4. MEAN. The MEAN of a set of data is the sum of the data divided by the number of items. Mean = sum of data - PowerPoint PPT Presentation
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CONFIDENTIAL 1
MEAN
CONFIDENTIAL 2
LET’S CHECK WHAT YOU KNOW
• Simplify1. 3+2+45+67+78+96 = 2912. (45+67+56+43+23)= 78 33. 9.261 = 3.087 34. 255 ÷ 4 = 63.755. 137 ÷ 5 = 27.4
CONFIDENTIAL 3
MEAN
• The MEAN of a set of data is the sum of the data divided by the number of items.
• Mean = sum of data number of items
CONFIDENTIAL 4
Example-1
• Find the mean of the following observations: 41, 80, 112, 86, 21
• Mean = (41+80+112+86+21) 5 = 340 = 68 5
CONFIDENTIAL 5
Mean
• A number that helps describe all of the data in a data set is an average or a measure of central tendency.
• One of the most common measures of central tendency is the mean.
CONFIDENTIAL 6
Outliers
• In statistics, a set of data may contain a value much higher or lower than the other values.
• This value is called an outlier.• Outliers can significantly
affect the mean.
CONFIDENTIAL 7
Example of Outliers
•Identify the outlier in the temperature data. Then find the mean with and without the outlier. Describe how the outlier affects the mean of the data.
Day Temp (°F)
Monday 80
Tuesday 81
Wednesday
40
Thursday 77
Friday 82
CONFIDENTIAL 8
Example of Outliers
•Compared to the other values, 40°F is extremely low. So, it is an outlier.•Mean with outlier
= 80+81+40+77+82
5
= 360
5
=72
•Mean without outlier = 80+81+77+82
4= 320 4=80
With the outlier, the mean is less than all but one of the data values.Without the outlier, the mean better represents the values in the data set
CONFIDENTIAL 9
Your Turn!
•Find the mean for the snowfall data with and without the outlier. Then tell how the outlier affects the mean of the data.
Month Snowfall (in.)
Nov. 20
Dec. 19
Jan. 20
Feb. 17
Mar. 4
CONFIDENTIAL 10
Answer
• Compared to the other values, 4 inches is low. So, it is an outlier.
• Mean with outlier = 20+19+20+17+4 =80 =165 5
• Mean without outlier = 20+19+20+17 =76 =19 4 4
With the outlier, the mean is less than the values of most of the data. Without the outlier, the mean is close in value to the data.
CONFIDENTIAL 11
Example-2
• Find the mean of the following data: 3, 10, 12, 7, 9, 8, 7.
• Mean = Sum of data Number of data = 3+10+12+7+9+8+7 7 = 56 = 8 7
CONFIDENTIAL 12
BREAK
CONFIDENTIAL 13
Do it by yourself• The marks obtained by 10
students in a test are: 90, 75, 80, 65, 78, 65, 80, 77, 86 and 74. Find the mean.
• Mean = 90+75+80+65+78+65+80+77+86+74
10 = 77
CONFIDENTIAL 14
Practice Problems
Find the mean of the following:
1. 72,55,48,61,63,37 = 562. 75,82,81,95,92 = 853. 7,15,9,12,10,13 = 114. 365,225,118,132,240 = 2165. 15,24,14,48,27,20= 24.67
CONFIDENTIAL 15
More problems• The marks Christy got in 6
tests are : 65 72, 59, 81, 68, 72. while another student Fredy wrote only 5 of these tests and his marks are 71, 54, 68, 82, 75. Whose overall performance is better?
CONFIDENTIAL 16
ANSWER• Mean marks of Christy = 65 + 72 + 59 + 81 + 68 + 72 6 = 417 = 69.5 6 • Mean marks of Fredy = 71 + 54 + 68 + 82 + 75 = 350 = 70
5 5Since Fredy has a higher mean, we can say that
Fredy’s performance is better.
CONFIDENTIAL 17
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Have Fun!
CONFIDENTIAL 18
Assignments
1. The minimum temperatures in Celsius recorded on various days of a week are 220, 250, 230, 210, 200, 270. Find their mean.
2. Calculate the mean weight of the students from the following data: 135, 184, 94, 288, 49, 200
CONFIDENTIAL 19
Answers
1. 22+25+23+21+20+27 = 138 = 23
6 6 2. 135+184+94+288+49+200 6 = 950 = 158.33 6
CONFIDENTIAL 20
Assignment 3. The maximum daily temperature (in
C0) for the first week of a month in Melbourne are 37, 35, 38, 34, 39, 36 and 33. calculate the mean temperature for the week.
Find the mean of :4. 3, 7, 8, 5, 6, 9, 6, 4, 65. 25, 31, 27, 29, 32, 25, 28, 25.6 3, 5, 1, 5, 7, 3, 2, 37 2, 3, 4, 4, 1, 6, 4, 3.
CONFIDENTIAL 21
Answers 3. Mean Temperature =
37+35+38+34+39+36+33 7 = 360C 4. 65. 27.75 6. 3.6257. 3.375
CONFIDENTIAL 22
Assignment 8. Which piece of data in the
data set 98, 103, 96, 147, 100, 85, 546, 120, 98 is an outlier? 546
9. Write a set of data in which the mean is affected by the outlier.
Sample Answer: 13, 13, 13, 15, 35, 15, 21
CONFIDENTIAL 23
Let’s Review what we studied today
• Mean of a set of data is the sum of the data divided by the number of items.
• Mean = sum of data number of items
CONFIDENTIAL 24
Good- ByeWe will meet you soon
