Find the mean, median, and mode of the following data Classes 0-10 10-20 20-30 30-40 40-50 50-60 60-70 Frequency 5 10 18 30 20 12 5

We solve this problem in three parts. (i) For finding the mean we create a table by adding the columns to given data. One column is for midpoint of class, xi and other column for product of frequency and midpoint of class fixi then we find mean using the formula x¯=∑fixi∑fi (ii) For finding the median we create a table by adding the columns to given data. One column is for cumulative frequency which means sum of frequencies up to that class interval. Then we find the median class for which class interval (N2)th frequency lies. Then median is given by median=L+⎛⎝⎜⎜N2−f0f⎞⎠⎟⎟h L is a lower interval of the median class N is the sum of all frequencies f0 is the cumulative frequency of preceding median class f is the frequency of the median class h is the height of class interval (iii) For calculating the mode we take the given data and find the modal class where the frequency is highest. Then, we formula of mode as Mode=L+(f1−f02f1−f0−f2)h L is lower interval of modal class f0 is frequency of preceding modal class f1 is frequency of modal class f2 is frequency of succeeding modal class h is height of class interval Complete step-by-step solution: (i) Let us solve for the mean. Let us create a table by adding the columns to given data. One column is for midpoint of class, xi and other column for product of frequency and midpoint of class fixi as follows- Class interval Frequency (fi ) Midpoint of class (xi ) fixi 0-10 5 5 25 10-20 10 15 150 20-30 18 25 450 30-40 30 35 1050 40-50 20 45 900 50-60 12 55 660 60-70 5 65 325 Now, we know that mean is calculated by using the formula x¯=∑fixi∑fi By substituting the required values in the formula we get ⇒x¯=25+150+450+1050+900+660+3255+10+18+30+20+12+5⇒x¯=3560100⇒x¯=35.6 Therefore, the mean of given data is 35.6. (ii) Let us solve for median Let us create a table by adding the columns to given data. One column is for cumulative frequency which means sum of frequencies up to that class interval as follows Class interval Frequency (fi ) Cumulative frequency 0-10 5 5 10-20 10 15 20-30 18 33 30-40 30 63 40-50 20 83 50-60 12 95 60-70 5 100 We know that N=∑fi=100 . Now, we need to find in which class (N2)th frequency lies. We know that N2=50 . So, frequency 50 lies in the class of 30. So, the median class is 30-40. We know that the formula of median is median=L+⎛⎝⎜⎜N2−f0f⎞⎠⎟⎟h L is lower interval of median class (L=30) N is sum of all frequencies (N=100) f0 is cumulative frequency of preceding median class (f0=33) f is frequency of median class (f=30) h is height of class interval (h=10) By substituting these values in the formula we get ⇒median=30+⎛⎝⎜⎜1002−3330⎞⎠⎟⎟×10⇒median=30+173⇒median=35.67 Therefore, the median of given data is 35.67. (iii) Let us solve for mode. Let us take the given data Class interval Frequency (fi ) 0-10 5 10-20 10 20-30 18 30-40 30 40-50 20 50-60 12 60-70 5 Here, we can see that the highest frequency is at 30-40. So, we can say that class 30-40 is the modal class. We know that the formula for mode as Mode=L+(f1−f02f1−f0−f2)h L is lower interval of modal class (L=30) f0 is frequency of preceding modal class (f0=18) f1 is frequency of modal class (f1=30) f2 is frequency of succeeding modal class (f2=20) h is height of class interval (h=10) By substituting the values in the formula we get ⇒Mode=30+(30−18(2×30)−18−20)×10⇒Mode=30+(1222)×10⇒Mode=35.45 Therefore, the mode of given data is 35.45. Note: The only difficulty where students will face is selecting the median and modal classes. After selecting the classes taking the frequencies and cumulative frequencies for substituting in the formula is important. Due to confusion students will take frequencies of median and modal classes instead of preceding and succeeding frequencies. The selection needs to be taken care of.