![Example 1: To find the median of 4,5,7,2,1 [ODD].
Step 1: Count the total numbers given.
There are 5 elements or numbers in the
distribution.
Step 2: Arrange the numbers in ascending order.
1,2,4,5,7
Step 3: The total elements in the distribution (5) is
odd.
The middle position can be calculated using the
formula. (n+1)/2
So the middle position is (5+1)/2 = 6/2 = 3 th Value
The number at 3rd position is = Median = 4](https://izqule7twkl7vq3ljkxejyz-s-a2157.bj.tsgdht.cn/biostatistics2-240617035932-897977f1/85/Biostatistics-community-medicine-or-Psm-pptx-16-320.jpg)
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- 2. • Measures of central tendency of a distribution - A numerical value that describes the central position of data • 3 common measures • Mean • Median • Mode
- 3. Arithmetic mean: It is obtained by summing up of all observations divided by number of observations. It is denoted by X (Sample Mean) Population mean is Denoted by µ (mu)
- 4. Mean (Arithmetic Mean) Mean (arithmetic mean) of data values Sample mean Population mean 1 1 2 n i i n X X X X X n n 1 1 2 N i i N X X X X N N Sample Size Size Population
- 5. • Measures of Location: Averages (Mean) Mean = Total or sum of the observations Number of observations = X1 + X2 + …. Xn = ΣX n n • The mean is calculated by different methods in two types of series, ungrouped and grouped series. 17/06/2024 Measures of Central tendancy 5
- 6. Ungrouped Series: In such series the number of observations is small and there are two methods for calculating the mean. The choice depends upon the size of observations in the series. (I). When the observations are small in size, simply add them up and divide by the number of observations. Example: 1. Tuberculin test reaction of 10 boys is arranged in ascending order being measured in millimeters. Find the mean size of reaction: 3,5,7,7,8,8,9,10,11,12 → Mean or = ΣX N = 3+5+7+7+8+8+9+10+11+12 10 = 80 10 =8 mm
- 7. Examples 2. Height in Centimeters for 7 school children are given below. • 148, 143, 160, 152,157, 150, 155 Cms. • Find the mean. • By direct method = ΣX = 1065 = 152.1 Centimeters n 7 17 June 2024 7 Mean, Median and Mode
- 8. By assumed mean (w) method X X-w=x (w=140) x 148 148-140 8 143 143-140 3 160 160-140 20 152 152-140 12 157 157-140 17 150 150-140 10 155 155-140 15 Σx= 85 x = Σ(X-w) = 85 =12.1 n 7 = w + x = 140 + 12.1 = 152.1Cm 17 June 2024 8 Mean, Median and Mode
- 9. Example. The average income of 10 lady doctors is Rs. 25000/- per month and that of 20 male doctors is Rs. 35000/- per month. Calculate the weighted mean or average income of all doctors. → For lady doctors X1 = Rs. 25000/- f1= 10 For male doctors X2 = Rs. 35000/- f2= 20 n = f1 + f2 = 10+20 =30 Total Income of lady doctors = X1 x f1 = ΣfX1 = 25000 x 10= 2,50,000 Total income of male doctors = X2 x f2 = ΣfX2 = 35000 x 20 =7,00,000 Total income of all doctors are= ΣfX = ΣfX1 + ΣfX2 =2,50,000 + 7,00,000= 9,50,000 The weighted mean income of all the doctors = ΣfX = 9,50,000 = Rs. 31,666.66 n 30 So, average income of all doctors is Rs. 31,666.66 Measures of Central tendancy 9
- 10. Example: Find the average weight of college students in kilogram from the table given below. Weight of students in Kg No. of students 60-<61 10 61- 20 62- 45 63- 50 64- 60 65- 40 66-<67 15 Total 240 17 June 2024 10 Mean, Median and Mode
- 11. 64- 60 65- 40 66-<67 15 Total 240 → 1st Method: Weight of students in Kg X Mid-point of each group Xg No. of students f fXg 60-<61 60.5 10 605 61- 61.5 20 1230 62- 62.5 45 2812.5 63- 63.5 50 3175 64- 64.5 60 3870 65- 65.5 40 2620 66-<67 66.5 15 997.5 Total n=240 ΣfXg=15310 Now, = ΣfXg = 15310 = 63.79Kg n 240 So, Mean weight of college students is 63.79Kg 17/06/2024 Measures of Central tendancy 11
- 12. 17/06/2024 Measures of Central tendancy 12 Weight of students in Kg X Mid-point of each group Xg No. of students f Working units x Groups weight fx Sum of fx 60-<61 60.5 10 -2 -20 61- 61.5 20 -1 -20 62- 62.5 (w) 45 0 0 -40 63- 63.5 50 +1 50 64- 64.5 60 +2 120 65- 65.5 40 +3 120 66-<67 66.5 15 +4 60 +350 Total n=240 Σfx=+310
- 13. 64- 64.5 60 +2 120 65- 65.5 40 +3 120 66-<67 66.5 15 +4 60 +350 Total n=240 Σfx=+310 Mean in working units = Σfx = 310 =1.29 n 240 Mean in real units = w + x Group interval = 62.5 + 1.29 = 63.79 Kg So, mean weight of college students is 63.79Kg 17/06/2024 Measures of Central tendancy 13
- 14. MEDIAN It is the value of middle observation after placing the observations in either ascending or descending order. Half the values lie above it and half below it.
- 15. UNGROUPED SERIES • If the number of observations is odd then median of the data will be n+1/2th observation • If even then median of the data will be the average of n/2th and ( n/2 ) +1th
- 16. Example 1: To find the median of 4,5,7,2,1 [ODD]. Step 1: Count the total numbers given. There are 5 elements or numbers in the distribution. Step 2: Arrange the numbers in ascending order. 1,2,4,5,7 Step 3: The total elements in the distribution (5) is odd. The middle position can be calculated using the formula. (n+1)/2 So the middle position is (5+1)/2 = 6/2 = 3 th Value The number at 3rd position is = Median = 4
- 17. Example 2 : To find the median of 5,7,2,1,6,4. step 1 : count the total numbers given. there are 6 numbers in the distribution. step 2 :arrange the numbers in ascending order. 1,2,4,5,6,7. step 3 :the total numbers in the distribution is 6 (even). so the average of two numbers which are respectively in positions n/2th and (n/2)+1th will be the median of the given data. Median = (4+5)/2 = 4.5
- 18. Mode • A measure of central tendency • Value that occurs most often • Not affected by extreme values • There may be no mode or several modes
- 19. • To find the mode of 11,3,5,11,7,3,11 • Arrange the numbers in ascending order. 3,3,5,7,11,11,11 Mode = 11
- 20. Measures of variability of individual observations: • i. Range • ii. Interquartile range • iii. Mean deviation • iv. Standard deviation • v. Coefficient of variation.
- 21. Measures of variability of samples: • i. Standard error of mean • ii Standard error of difference between two means • iii Standard error of proportion • iv Standard error of difference between two proportions • v. Standard error of correlation coefficient • vi. Standard deviation of regression coefficient.
- 22. 22 The Range • The range is defined as the difference between the largest score in the set of data and the smallest score in the set of data, • XL - XS • What is the range of the following data: 4 8 1 6 6 2 9 3 6 9 • The largest score (XL) is 9; the smallest score (XS) is 1; the range is XL - XS = 9 - 1 = 8
- 23. Quartiles •Split Ordered Data into 4 Quarters • • Position of i-th Quartile: position of point 25% 25% 25% 25% Q1 Q2 Q3 Q i(n+1) i 4 Data in Ordered Array: 11 12 13 16 16 17 18 21 22 Position of Q1 = 2.50 Q1 =12.5 = 1•(9 + 1) 4
- 24. • Measure of Variation • Also Known as Midspread: Spread in the Middle 50% • Difference Between Third & First Quartiles: Interquartile Range = • Not Affected by Extreme Values Interquartile Range 1 3 Q Q Data in Ordered Array: 11 12 13 16 16 17 18 21 22 1 3 Q Q = 17.5 - 12.5 = 5
- 25. Box-and-Whisker Plot • Graphical Display of Data Using 5-Number Summary Median 4 6 8 10 12 Q3 Q1 Xlargest Xsmallest
- 26. 3. Mean Deviation (M.D.): MD = X−𝑋 𝑛
- 27. • 𝑋 = 𝑋 𝑛 = 775 5 = 155 cm • Mean deviation MD = X−𝑋 𝑛 = 40 5 = 8 cm • Mean deviation is not used in statistical analysis being less mathematical value, particularly in drawing inferences. Observations (X) X − 𝑋 X − 𝑋 150 -5 5 160 +5 5 155 0 0 170 +15 15 140 -15 15 ΣX= 775 Σ X − 𝑋 = 40 e.g. Height of 5 students in Centimeter
- 28. • Root-mean squared deviation called SD. SD = X−𝑋 2 𝑛−1 When sample size is less than 30 • The formula becomes • SD = X−𝑋 2 𝑛 • When sample size is more than 30
- 29. Observation X Deviation from Mean x= X- Square of deviation x2 =( X - )2 23 +3 9 22 +2 4 20 0 0 24 +4 16 16 -4 16 17 -3 9 18 -2 4 19 -1 1 21 +1 1 ΣX=180 0 Σ=( X - )2=60 Calculation of Standard Deviations in Ungrouped series: Example: Find the mean respiratory rate per minute and its SD when in 9 cases the rate was found to be 23, 22, 20, 24, 16, 17, 18, 19 and 21. = ΣX = 180 = 20/minute N 9 So s or SD= = =2.74 min Σ(X - )2 n -1 60/9-1
- 30. 30 Standard Deviation • Standard deviation = variance • Variance = standard deviation2
- 31. • It is a measure used to compare relative variability 5. Coefficientof Variation:
- 32. Persons Mean Ht in Cm SD in Cm Adults 160cm 10cm Children 60cm 5cm In two series of adults aged 21 years and children 3 months old following values were obtained for height. Find which series shows greater variation? CV = SD x 100 Mean CV of adults = 10 x 100 = 6.25% 160 CV of children = 5 x 100 = 8.33% 60 Thus it is found that heights in children show greater variation than in adults.
- 35. (a) The area between one standard deviation( SD) on either side of the mean ( x ± l ϭ ) will include approximately 68% of the values in the distribution (b) The area between two standard deviations on either side of the mean ( x ± 2 ϭ ) will cover most of the values, i.e., approximately 95 % of the values (c) The area between three standard deviations on either side of the mean ( x ± 3 ϭ) will include 99.7 % of the values. • These limits on either side of the mean are called “confidence limits
- 36. Properties of STANDARD NORMAL CURVE • Bell shaped & smooth curve • Two tailed & symmetrical • Tail doesn’t touch the base line • Area under the curve is 1 • Mean = 0 • Standard deviation is =1 • Mean, Median, Mode coincide • Two inflection- Convex at centre, convert to Concave while descending to periphery • Perpendicular drawn from the point of inflection cut the base at 1 standard deviation • Approximately 68%, 95%, 99% observations are included in the range of Mean +/- 1SD, 2 SD, 3 SD respectively • No portion of the curve lie below the X axis -5 -4 -3 -2 -1 0 1 2 3 4 5 68.26% 95.44% 99.72%
- 37. Example Q: 95% of students at school are between 1.1m and 1.7m tall. Assuming this data is normally distributed calculate the mean and standard deviation?
- 38. 95% is 2 standard deviations either side of the mean (a total of 4 standard deviations) so: 1 standard deviation = (1.7m-1.1m) / 4 = 0.6m / 4 = 0.15m And this is the result: The mean is halfway between 1.1m and 1.7m: Mean = (1.1m + 1.7m) / 2 = 1.4m
- 39. Q: 68% of the marks in a test are between 51 and 64, Assuming this data is normally distributed, what are the mean and standard deviation? Example
- 40. Answer: The mean is halfway between 51 and 64: Mean = (51 + 64)/2 = 57.5 68% is 1 standard deviation either side of the mean (a total of 2 standard deviations) so: 1 standard deviation = (64 - 51)/2 = 13/2 = 6.5
- 41. Example Q: Average weight of baby at birth is 3.05 kg with SD of 0.39kg. If the birth weights are normally distributed would you regard: 1.Weight of 4 kg as abnormal? 2. Weight of 2.5 kg as normal?
- 42. • Answer Normal limits of weight will be within range of Mean + 2 SD = 3.05 + (2 x 0.39) = 3.05 + 0.78 = 2.27 to 3.83 1. The wt of 4 kg lies outside the normal limits. So it is taken as abnormal. 2. The wt. of 2.5 kg lies within the normal limits. So it is taken as normal.
- 43. Standard normal deviate The distance of a value (x) from the mean (X bar) of the curve in units of standard deviation is called “relative deviate or standard normal deviate” and usually denoted by Z. Z = Observation –Mean Standard Deviation
- 44. Q: The pulse of a group of normal healthy males was 72, with a standard deviation of 2. What is the probability that a male chosen at random would be found to have a pulse of 80 or more ?
- 45. THANK YOU