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MEAN

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

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MEAN

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  1. MEAN CONFIDENTIAL

  2. 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 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. CONFIDENTIAL

  8. 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 Example of Outliers 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. 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 =16 5 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: • 72,55,48,61,63,37 = 56 • 75,82,81,95,92 = 85 • 7,15,9,12,10,13 = 11 • 365,225,118,132,240 = 216 • 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 5 Since Fredy has a higher mean, we can say that Fredy’s performance is better. CONFIDENTIAL

  17. Have Fun! http://www.thekidzpage.com/freekidsgames/games/breakdown/index.html CONFIDENTIAL

  18. Assignments • The minimum temperatures in Celsius recorded on various days of a week are 220, 250, 230, 210, 200, 270. Find their mean. • Calculate the mean weight of the students from the following data: 135, 184, 94, 288, 49, 200 CONFIDENTIAL

  19. Answers • 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, 6 5. 25, 31, 27, 29, 32, 25, 28, 25. • 3, 5, 1, 5, 7, 3, 2, 3 • 2, 3, 4, 4, 1, 6, 4, 3. CONFIDENTIAL

  21. Answers 3. Mean Temperature = 37+35+38+34+39+36+33 7 = 360C • 6 • 27.75 • 3.625 • 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 CONFIDENTIAL

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