# Class 11 RD Sharma Solutions – Chapter 32 Statistics – Exercise 32.3

**Question 1.** **Compute the mean deviation from the median of the following distribution:**

Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |

Frequency | 5 | 10 | 20 | 5 | 10 |

**Solution:**

Calculating the median:

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free classeswhich will definitely help them in making a wise career choice in the future.Median is the middle term of the observation in ascending order, X

_{i},Here, the middle term is 25.

Class Interval x _{i}f _{i}Cumulative Frequency |d _{i}| = |x_{i}– M|f _{i}|d_{i}|0-10 5 5 5 20 100 10-20 15 10 15 10 100 20-30 25 20 35 0 0 30-40 35 5 91 10 50 40-50 45 10 101 20 200 Total = 50 Total = 450 Therefore, Median = 25

Mean Deviation,

MD= 1/50 × 450

= 9

Therefore, mean deviation is 9.

**Question 2. Find the mean deviation from the mean for the following data:**

**(i)**

Classes | 0-100 | 100-200 | 200-300 | 300-400 | 400-500 | 500-600 | 600-700 | 700-800 |

Frequencies | 4 | 8 | 9 | 10 | 7 | 5 | 4 | 3 |

**(ii)**

Classes | 95-105 | 105-115 | 115-125 | 125-135 | 135-145 | 145-155 |

Frequencies | 9 | 13 | 16 | 26 | 30 | 12 |

**Solution:**

(i)Mean = 17900/50

= 358

Also, the number of observations, N=50

Class Interval x _{i}f _{i}Cumulative Frequency |d _{i}| = |x_{i}– M|f _{i}|d_{i}|0-100 50 4 200 308 1232 100-200 150 8 1200 208 1664 200-300 250 9 2250 108 972 300-400 350 10 3500 8 80 400-500 450 7 3150 92 644 500-600 550 5 2750 192 960 600-700 650 4 2600 292 1168 700-800 750 3 2250 392 1176 Total = 50 Total = 17900 Total = 7896 = 1/50 × 7896

= 157.92

Therefore, mean deviation is 157.92.

(ii)Mean = 13630/106

= 128.58

Also, the number of observations, N=106

Class Interval x _{i}f _{i}Cumulative Frequency |d _{i}| = |x_{i}– M|f _{i}|d_{i}|95-105 100 9 900 28.58 257.22 105-115 110 13 1430 18.58 241.54 115-125 120 16 1920 8.58 137.28 125-135 130 26 3380 1.42 36.92 135-145 140 30 4200 11.42 342.6 145-155 150 12 1800 21.42 257.04 N = 106 Total = 13630 Total = 1272.6 = 1/106 × 1272.6

= 12.005

**Question 3.** **Compute mean deviation from mean of the following distribution:**

Marks | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 |

No. of students | 8 | 10 | 15 | 25 | 20 | 18 | 9 | 5 |

**Solution:**

= 5390/110

= 49

Class Interval x _{i}f _{i}Cumulative Frequency |d _{i}| = |x_{i}– M|f _{i}|d_{i}|10-20 15 8 120 34 272 20-30 25 10 250 24 240 30-40 35 15 525 14 210 40-50 45 25 1125 4 100 50-60 55 20 1100 6 120 60-70 65 18 1170 16 288 70-80 75 9 675 26 234 80-90 85 5 425 36 180 N = 110 Total = 5390 Total = 1644 = 1/110 × 1644

= 14.94

Therefore, mean deviation is 14.94.

**Question 4.** **The age distribution of 100 life-insurance policyholders is as follows:**

Age (on nearest birthday) | 17-19.5 | 20-25.5 | 26-35.5 | 36-40.5 | 41-50.5 | 51-55.5 | 56-60.5 | 61-70.5 |

No. of persons | 5 | 16 | 12 | 26 | 14 | 12 | 6 | 5 |

**Calculate the mean deviation from the median age.**

**Solution:**

Number of observations, N = 96

So, N/2 = 96/2 = 48

The cumulative frequency just greater than 48 is 59, and therefore, the corresponding value of x is 38.25

Hence, Median = 38.25

Class Interval x _{i}f _{i}Cumulative Frequency |d _{i}| = |x_{i}– M|f _{i}|d_{i}|17-19.5 18.25 5 5 20 100 20-25.5 22.75 16 21 15.5 248 36-35.5 30.75 12 33 7.5 90 36-40.5 38.25 26 59 0 0 41-50.5 45.75 14 73 7.5 105 51-55.5 53.25 12 85 15 180 56-60.5 58.25 6 91 20 120 61-70.5 65.75 5 96 27.5 137.5 Total = 96 Total = 980.5 Number of observations, N = 96.

= 1/96 × 980.5

= 10.21

∴ The mean deviation is 10.21

**Question 5.** **Find the mean deviation from the mean and from a median of the following distribution:**

Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |

No. of students | 5 | 8 | 15 | 16 | 6 |

**Solution:**

Number of observations, N = 50

So, N/2 = 50/2 = 25

The cumulative frequency just greater than 25 is 58, and the corresponding value of x is 28.

Therefore, Median = 28

Now,

= 1350/50

= 27

Class Interval x _{i}f _{i}Cumulative Frequency |d _{i}| = |x_{i}– Median|f _{i}|d_{i}|F _{i}X_{i}|X _{i}– Mean|F _{i }|X_{i}– Mean|0-10 5 5 5 23 115 25 22 110 10-20 15 8 13 13 104 120 12 96 20-30 25 15 28 3 45 375 2 30 30-40 35 16 44 7 112 560 8 128 40-50 45 6 50 17 102 270 18 108 N = 50 Total = 478 Total = 1350 Total = 472 Mean deviation from the median of observation = 478/50 = 9.56

And, Mean deviation from mean of observation = 472/50 = 9.44

∴ The mean deviation from the median is 9.56 and from the mean is 9.44.

**Question ****6. Calculate mean deviation about median age for the age distribution of 100 persons given below:**

Age | 16-20 | 21-25 | 26-30 | 31-35 | 36-40 | 41-45 | 45-50 | 50-55 |

Number of persons | 5 | 6 | 12 | 14 | 26 | 12 | 16 | 9 |

**Solution:**

Converting the given data into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.

Age x _{i}f _{i}Cumulative Frequency |d _{i}| = |x_{i}–38|f _{i}|d_{i}|15.5-20.5 18 5 5 20 100 20.5-25.5 23 6 11 15 90 25.5-30.5 28 12 23 10 120 30.5-35.5 33 14 37 5 70 35.5-40.5 38 26 63 0 0 40.5-45.5 43 12 75 5 60 45.5-50.5 48 16 91 10 160 50.5-55.5 53 9 100 15 135 N = 100 Total = 735 We have, N = 100

So, N/2 = 100/2 = 50

The cumulative frequency just greater than N/2 is 63 and the corresponding class is 35.5-40.5.

l=35.5, f=26, h= 5, F =37

Therefore, Median = l + (N/2 – F)/f * h = 35.5 + 50-37/26 * 5 =38

**Question 7. Calculate the mean deviation from the median of the following data:**

Class interval | 0-4 | 4-8 | 8-12 | 12-16 | 16-20 |

Frequency | 4 | 6 | 8 | 5 | 2 |

**Solution:**

f _{i}x _{i}f _{i}x_{i}|x _{i}-9.2|f _{i}|x_{i}-9.2|0-4 4 2 8 7.2 28.8 4-8 6 6 36 3.2 19.2 8-12 8 10 80 0.8 6.4 12-16 5 14 70 4.8 24.0 16-20 2 18 36 8.8 17.6 N=25 Total=230 total = 96.0 Mean = 230/25 =9.2

Mean Deviation=96/25 = 3.84

**Question 8. Calculate the mean deviation from the median of the following data:**

Class interval | 0-6 | 6-12 | 12-18 | 18-24 | 24-30 |

Frequency | 4 | 5 | 3 | 6 | 2 |

**Solution:**

f _{i}x _{i}f _{i}x_{i}|x _{i}-14.1|f _{i}|x_{i}-14.1|0-6 4 3 12 11.1 44.4 6-12 5 9 45 5.1 25.5 12-18 3 15 45 0.9 2.7 18-24 6 21 126 6.9 41.4 24-30 2 27 54 12.9 25.8 N=20 Total=282 total = 139.8 Mean = 282/20 =14.1

Mean Deviation= 139.8/20 = 6.99