# Class 10 RD Sharma Solution – Chapter 7 Statistics – Exercise 7.4 | Set 1

**Question 1. Following are the lives in hours of 15 pieces of the components of aircraft engine. Find the median:**

**715, 724, 725, 710, 729, 745, 694, 699, 696, 712, 734, 728, 716, 705, 719.**

**Solution:**

On arranging the observations in ascending order, we have

694, 696, 699, 705, 710, 712, 715, 716, 719, 721, 725, 728, 729, 734, 745

Number of terms in the observation sequence is odd, i.e., N = 15

Now,

Median = (N + 1)/2

^{th }term= (15 + 1)/2

^{th}term= 8

^{th}termTherefore, 716, which is the 8

^{th}term is the median of the data.

**Question 2. The following is the distribution of height of students of a certain class in a certain city:**

Height (in cm): | 160 – 162 | 163 – 165 | 166 – 168 | 169 – 171 | 172 – 174 |

No of students: | 15 | 118 | 142 | 127 | 18 |

**Find the median height.**

**Solution:**

Class interval (exclusive) Class interval (inclusive) Class interval frequency Cumulative frequency 160 – 162 159.5 – 162.5 15 15 163 – 165 162.5 – 165.5 118 133(F) 166 – 168 165.5 – 168.5 142(f) 275 169 – 171 168.5 – 171.5 127 402 172 – 174 171.5 – 174.5 18 420 N =420 We have N = 420,

So, N/2 = 420/ 2 = 210

Now, The cumulative frequency just greater than N/2 is 275

Therefore, 165.5 – 168.5 is the median class s.t

L = 165.5, f = 142, F = 133 and h = (168.5 – 165.5) = 3

= 165.5 + 1.63

= 167.13

**Question 3. Following is the distribution of I.Q of 100 students. Find the median I.Q.**

I.Q: | 55 – 64 | 65 – 74 | 75 – 84 | 85 – 94 | 95 – 104 | 105 – 114 | 115 – 124 | 125 – 134 | 135 – 144 |

No of students: | 1 | 2 | 9 | 22 | 33 | 22 | 8 | 2 | 1 |

**Solution:**

Class interval (exclusive) Class interval (inclusive) Class interval frequency Cumulative frequency 55 – 64 54.5 – 64.5 1 1 65 – 74 64.5 – 74.5 2 3 75 – 84 74.5 – 84.5 9 12 85 – 94 84.5 – 94.5 22 34(F) 95 – 104 94.5 – 104.5 33(f) 67 105 – 114 104.5 – 114.5 22 89 115 – 124 114.5 – 124.5 8 97 125 – 134 124.5 – 134.5 2 98 135 – 144 134.5 – 144.5 1 100 N = 100 N = 100,

Therefore, N/2 = 100/ 2 = 50

The cumulative frequency just greater than N/ 2 is 67 then the median class is (94.5 – 104.5) s.t,

L = 94.5, F = 33, h = (104.5 – 94.5) = 10

= 94.5 + 4.85

= 99.35

**Question 4. Calculate the median from the following data:**

Rent (in Rs): | 15 – 25 | 25-35 | 35-45 | 45-55 | 55-65 | 65-75 | 75-85 | 85-95 |

No of houses: | 8 | 10 | 15 | 25 | 40 | 20 | 15 | 7 |

**Solution:**

Class intervalFrequencyCumulative frequency15 – 25 8 8 25 – 35 10 18 35 – 45 15 33 45 – 55 25 58 (F) 55 – 65 40(f) 98 65 – 75 20 118 75 – 85 15 133 85 – 95 7 140 N = 140 N = 140,

And, N/2 = 140/ 2 = 70

The cumulative frequency just greater than N/ 2 is 98 then median class is 55 – 65 s.t,

L = 55, f = 40, F = 58, h = 65 – 55 = 10

**Question 5. Calculate the median from the following data:**

Marks below: | 0-10 | 10 – 20 | 20 – 30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |

No of students: | 15 | 35 | 60 | 84 | 96 | 127 | 198 | 250 |

**Solution:**

Marks below No. of students Class interval Frequency Cumulative frequency 10 15 0-10 15 15 20 35 10-20 20 35 30 60 20-30 25 60 40 84 30-40 24 84 50 96 40-50 12 96(F) 60 127 50-60 31(f) 127 70 198 60-70 71 198 80 250 70-80 52 250 N = 250 N = 250,

And, N/2 = 250/ 2 = 125

The cumulative frequency just greater than N/ 2 is 127 then median class is 50 – 60 s.t,

L = 50, f = 31, F = 96, h = 60 -50 = 10

**Question 6. Calculate the missing frequency from the following distribution, it being given that the median of the distribution is 24.**

Age in years: | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |

No of persons: | 5 | 25 | ? | 18 | 7 |

**Solution:**

Let us assume the unknown frequency to be x.

Class interval Frequency Cumulative frequency 0 – 10 5 5 10-20 25 30 (F) 20-30 x(f) 30 + x 30-40 18 48 + x 40-50 7 55 + x N=170 Given: Median = 24

Therefore,

Median class = 20 – 30; L = 20, h = 30 -20 = 10, f = x, F = 30

4x = 275 + 5x – 300

4x – 5x = – 25

– x = – 25

x = 25

Therefore, x = 25

**Question 7. The following table gives the frequency distribution of married women by age at marriage.**

Age (in years) | Frequency | Age (in years) | Frequency |

15 – 19 | 53 | 40 – 44 | 9 |

20 – 24 | 140 | 45 – 49 | 5 |

25 – 29 | 98 | 45 – 49 | 3 |

30 – 34 | 32 | 55 – 59 | 3 |

35 – 39 | 12 | 60 and above | 2 |

**Calculate the median and interpret the results.**

**Solution:**

Class interval (exclusive)Class interval (inclusive)FrequencyCumulative frequency15 – 19 14.5 – 19.5 53 53(F) 20 – 24 19.5 – 24.5 140(f) 193 25 – 29 24.5 – 29.5 98 291 30 – 34 29.5 – 34.5 32 323 35 – 39 34.5 – 39.5 12 335 40 – 44 39.5 – 44.5 9 344 45 – 49 44.5 – 49.5 5 349 50 – 54 49.5 – 54.5 3 352 55 – 54 54.5 – 59.5 3 355 60 and above 59.5 and above 2 357 N =357 N = 357,

And, N/2 = 357/ 2 = 178.5

The cumulative frequency just greater than N/2 is 193,

Therefore, median class is 19.5 – 24.5 s.t

l = 19.5, f = 140, F = 53, h = 25.5 – 19.5 = 5

Median = 23.98, that implies that nearly half of the women are married between the ages of 15 and 25.

**Question 8. The following table gives the distribution of the life time of 400 neon lamps:**

Life time: (in hours) | Number of lamps |

1500 – 2000 | 14 |

2000 – 2500 | 56 |

2500 – 3000 | 60 |

3000 – 3500 | 86 |

3500 – 4000 | 74 |

4000 – 4500 | 62 |

4500 – 5000 | 48 |

**Find the median life.**

**Solution:**

Life timeNumber of lamps fiCumulative frequency (cf)1500 – 2000 14 14 2000 – 2500 56 70 2500 – 3000 60 130(F) 3000 – 3500 86 216 3500 – 4000 74 290 4000 – 4500 62 352 4500 – 5000 48 400 N = 400 Now

N = 400

And the cumulative frequency just greater than n/2 (= 200) is 216, which belongs to the class interval 3000 – 3500

Median class = 3000 – 3500. Therefore,

(l) = 3000 and,(f) of median class = 86, (cf) of class preceding median class = 130 and (h) = 500

We have,

= 3000 + (35000/86)

= 3406.98 hrs, which is the median time of lamps.

**Question 9. The distribution below gives the weight of 30 students in a class. Find the median weight of students:**

Weight (in kg): | 40 – 45 | 45 – 50 | 50 – 55 | 55 – 60 | 60 – 65 | 65 – 70 | 70 – 75 |

No of students: | 2 | 3 | 8 | 6 | 6 | 3 | 2 |

**Solution:**

Weight (in kg) Number of students fi Cumulative frequency (cf) 40 – 45 2 2 45-50 3 5 50-55 8 13 55-60 6 19 60-65 6 25 65-70 3 28 70-75 2 30 The cf value just greater than n/ 2 (i.e. 30/ 2 = 15) is 19, belongs to class interval 55 – 60.

Therefore,

Median class = 55 – 60

where,

(l) of median class = 55, (f) of median class = 6, (cf) = 13 and (h) = 5

We have,

= 55 + 10/6 = 56.666 which is approximately 56.67 kg.

**Question 10. Find the missing frequencies and the median for the following distribution if the mean is 1.46**

No. of accidents: | 0 | 1 | 2 | 3 | 4 | 5 | Total |

Frequencies (no. of days): | 46 | ? | ? | 25 | 10 | 5 | 200 |

**Solution:**

No. of accidents (x)No. of days (f)fx0 46 0 1 x x 2 y 2y 3 25 75 4 10 40 5 5 25 N = 200 Sum = x + 2y + 140 Since, we know,

N = 200

Substituting values, we get,

⇒ 46 + x + y + 25 + 10 + 5 = 200

⇒ x + y = 200 – 46 – 25 – 10 – 5

⇒ x + y = 114…… (i)

Also, Mean = 1.46

⇒ Sum/ N = 1.46

Substituting values,

⇒ (x + 2y + 140)/ 200 = 1.46

⇒ x + 2y = 292 – 140

⇒ x + 2y = 152 …….(ii)

Solving from (i) and (ii), we get

x + 2y – x – y = 152 – 114

⇒ y = 38

And, x = 114 – 38 = 76 (from equation (i))

Now, putting the values, we get,

N = 200 N/2 = 200/2 = 100

So, the cumulative frequency just greater than N/2 is 122

And, therefore, the median is 1.

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