# Class 10 RD Sharma Solutions – Chapter 7 Statistics – Exercise 7.3 | Set 2

**Question 14. Find the mean of the following frequency distribution:**

Class interval: | 25 – 29 | 30 – 34 | 35 – 39 | 40 – 44 | 45 – 49 | 50 – 54 | 55 – 59 |

Frequency: | 14 | 22 | 16 | 6 | 5 | 3 | 4 |

**Solution:**

Let’s consider the assumed mean (A) = 42

Class intervalMid – value x_{i}d_{i }= x_{i }– 42u_{i }= (x_{i }– 42)/5f_{i}f_{i}u_{i}25 – 29 27 -15 -3 14 -42 30 – 34 32 -10 -2 22 -44 35 – 39 37 -5 -1 16 -16 40 – 44 42 0 0 6 0 45 – 49 47 5 1 5 5 50 – 54 52 10 2 3 6 55 – 59 57 15 3 4 12 N = 70Σ f_{i}u_{i }= -79From the table it’s seen that,

A = 42 and h = 5

Mean = A + h x (Σf

_{i }u_{i}/N)= 42 + 5 x (-79/70)

= 42 – 79/14

= 42 – 5.643

= 36.357

**Question 15. For the following distribution, calculate mean using all suitable methods:**

Size of item: | 1 – 4 | 4 – 9 | 9 – 16 | 16 – 20 |

Frequency: | 6 | 12 | 26 | 20 |

**Solution:**

By direct method

Class intervalMid value x_{i}Frequency f_{i}f_{i}x_{i}1 – 4 2.5 6 15 4 – 9 6.5 12 18 9 – 16 12.5 26 325 16 – 27 21.5 20 430 N = 64Sum = 848Mean = (sum/N) + A

= 848/64

= 13.25

By assuming mean method

Let the assumed mean (A) = 65

Class intervalMid value x_{i}u_{i}= (x_{i}– A) = x_{i}– 65Frequency f_{i}f_{i}u_{i}1 – 4 2.5 -4 6 -25 4 – 9 6.5 0 12 0 9 – 16 12.5 6 26 196 16 – 27 21.5 15 20 300 N = 64Sum = 432Mean = A + sum/N

= 6.5 + 6.75

= 13.25

**Question 16. The weekly observation on cost of living index in a certain city for the year 2004 – 2005 are given below. Compute the weekly cost if living index.**

Cost of living index | Number of students | Cost of living index | Number of students |

1400 – 1500 | 5 | 1700 – 1800 | 9 |

1500 – 1600 | 10 | 1800 – 1900 | 6 |

1600 – 1700 | 20 | 1900 – 2000 | 2 |

**Solution:**

Let the assumed mean (A) = 1650

Class intervalMid value x_{i}d_{i}= x_{i}– A = x_{i}– 1650Frequency f_{i}f_{i}u_{i}1400 – 1500 1450 -200 -2 5 -10 1500 – 1600 1550 -100 -1 10 -10 1600 – 1700 1650 0 0 20 0 1700 – 1800 1750 100 1 9 9 1800 – 1900 1850 200 2 6 12 1900 – 2000 1950 300 3 2 6 N = 52Sum = 7We have

A = 16, h = 100

Mean = A + h (sum/N)

= 1650 + (175/13)

= 21625/13

= 1663.46

**Question 17. The following table shows the marks scored by 140 students in an examination of a certain paper:**

Marks: | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |

Number of students: | 20 | 24 | 40 | 36 | 20 |

**Calculate the average marks by using all the three methods: direct method, assumed mean deviation and shortcut method.**

**Solution:**

(i) Direct method:

Class intervalMid value x_{i}Frequency f_{i}f_{i}x_{i}0 – 10 5 20 100 10 – 20 15 24 360 20 – 30 25 40 1000 30 – 40 35 36 1260 40 – 50 45 20 900 N = 140Sum = 3620Mean = sum/N

= 3620/140

= 25.857

(ii) Assumed mean method:Let the assumed mean = 25

Mean = A + (sum/N)

Class intervalMid value x_{i}u_{i}= (x_{i}– A)Frequency f_{i}f_{i}u_{i}0 – 10 5 -20 20 -400 10 – 20 15 -10 24 -240 20 – 30 25 0 40 0 30 – 40 35 10 36 360 40 – 50 45 20 20 400 N = 140Sum = 120Mean = A + (sum/N)

= 25 + (120/140)

= 25 + 0.857

= 25.857

(iii) Step deviation method:Let the assumed mean (A) = 25

Class intervalMid value x_{i}d_{i}= x_{i}– A = x_{i}– 25Frequency f_{i}f_{i}u_{i}0 – 10 5 -20 -2 20 -40 10 – 20 15 -10 -1 24 -24 20 – 30 25 0 0 40 0 30 – 40 35 10 1 36 36 40 – 50 45 20 2 20 40 N = 140Sum = 12Mean = A + h(sum/N)

= 25 + 10(12/140)

= 25 + 0.857

= 25.857

**Question 18. The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the miss frequency f**_{1} and f_{2}.

_{1}and f

_{2}.

Class: | 0 – 20 | 20 – 40 | 40 – 60 | 60 – 80 | 80 – 100 | 100 – 120 |

Frequency: | 5 | f_{1} | 10 | f_{2} | 7 | 8 |

**Solution:**

Class intervalMid value x_{i}Frequency f_{i}f_{i}x_{i}0 – 20 10 5 50 20 – 40 30 f _{i}30f _{i}40 – 60 50 10 500 60 – 80 70 f _{2}70f _{2}80 – 100 90 7 630 100 – 120 110 8 880 N = 50Sum = 30f_{1}+ 70f_{2}+ 2060Given,

Sum of frequency = 50

5 + f

_{1}+ 10 + f_{2}+ 7 + 8 = 50f

_{1}+ f_{2}= 203f

_{1}+ 3f_{2}= 60 —(1) [Multiply both side by 3]and mean = 62.8

Sum/N = 62.8

(30f

_{1}+ 70f_{2}+ 2060)/50 = 62.830f

_{1}+ 70f_{2}= 3140 – 206030f

_{1}+ 70f_{2}= 10803f

_{1}+ 7f_{2}= 108 —(2) [divide it by 10]Subtract equation (1) from equation (2)

3f

_{1}+ 7f_{2}– 3f_{1}– 3f_{2}= 108 – 604f

_{2}= 48f

_{2}= 12Put value of f

_{2}in equation (1)3f

_{1}+ 3(12) = 60f

_{1}= 24/3 = 8f

_{1}= 8, f_{2}= 12

**Question 19. The following distribution shows the daily pocket allowance given to the children of a multistory building. The average pocket allowance is Rs 18.00. Find out the missing frequency.**

Class interval: | 11 – 13 | 13 – 15 | 15 – 17 | 17 – 19 | 19 – 21 | 21 – 23 | 23 – 25 |

Frequency: | 7 | 6 | 9 | 13 | – | 5 | 4 |

**Solution:**

Given mean = 18,

Let the missing frequency be v

Class intervalMid interval x_{i}Frequency f_{i}f_{i}x_{i}11 – 13 12 7 84 13 – 15 14 6 88 15 – 17 16 9 144 17 – 19 18 13 234 19 – 21 20 x 20x 21 – 23 22 5 110 23 – 25 14 4 56 N = 44 + xSum = 752 + 20xMean = sum/N

18 = 752 + 20×44 + x

792 + 18x = 752 + 20x

2x = 40

x = 20

**Question 20. If the mean of the following distribution is 27. Find the value of p.**

Class: | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |

Frequency: | 8 | P | 12 | 13 | 10 |

**Solution:**

Class intervalMid value x_{i}Frequency f_{i}f_{i}x_{i}0 – 10 5 8 40 10 – 20 15 P 152 20 – 30 25 12 300 30 – 40 35 13 455 40 – 50 45 16 450 N = 43 + PSum = 1245 + 15pGiven mean = 27

Mean = sum/N

1245 + 15p43 + p = 27

1245 + 15p = 1161 + 27p

12p = 84

P =7

**Question 21. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contain varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.**

Number of mangoes: | 50 – 52 | 53 – 55 | 56 – 58 | 59 – 61 | 62 – 64 |

Number of boxes: | 15 | 110 | 135 | 115 | 25 |

**Find the mean number of mangoes kept in packing box. Which method of finding the mean did you choose?**

**Solution:**

Number of mangoesNumber of boxes50 – 52 15 53 – 55 110 56 – 58 135 59 – 61 115 62 – 64 25 We may observe that class internals are not continuous

There is a gap between two class intervals. So we have to add ½ from lower class limit to each interval and class mark (x

_{i}) may be obtained by using the relationx

_{i}= upperlimit + lowerclasslimit2Class size (h) of this data = 3

Now taking 57 as assumed mean (a) we may calculated d

_{i}, u_{i}, f_{i}u_{i}as follows

Class intervalFrequency f_{i}Mid values x_{i}d_{i}= x_{i}– A = x_{i}– 25f_{i}u_{i}49.5 – 52.5 15 51 -6 -2 -30 52.5 – 55.5 110 54 -3 -1 -110 55.5 – 58.5 135 57 0 0 0 58.5 – 61.5 115 60 3 1 115 61.5 – 64.5 25 63 6 2 50 TotalN = 400Sum = 25Now we have N

Sum = 25

Mean = A +h (sum/N)

= 57 + 3 (45/400)

= 57 + 3/16

= 57 + 0.1875

= 57.19

Clearly mean number of mangoes kept in packing box is 57.19

**Question 22. The table below shows the daily expenditure on food of 25 households in a locality**

Daily expenditure (In Rs): | 100 – 150 | 150 – 200 | 200 – 250 | 250 – 300 | 300 – 350 |

Number of households: | 4 | 5 | 12 | 2 | 2 |

**Find the mean daily expenditure on food by a suitable method.**

**Solution:**

We may calculate class mark (x

_{i}) for each interval by using the relationx

_{i}= upperlimit + lowerclasslimit2Class size = 50

Now, Taking 225 as assumed mean (x

_{i}) we may calculate d_{i}, u_{i}, f_{i}u_{i}as follows

Daily expenditureFrequency f_{1}Mid value x_{i}d_{i}= x_{i}– 225f_{i}u_{i}100 – 150 4 125 -100 -2 -8 150 – 200 5 175 -50 -1 -5 200 – 250 12 225 0 0 0 250 – 300 2 275 50 1 2 300 – 350 2 325 100 2 4 N = 25Sum = -7Now we may observe that

N = 25

Sum = -7

225 + 50 (-7/25)

225 – 14 = 211

So, mean daily expenditure on food is Rs 211

**Question 23. To find out the concentration of SO**_{2} in the air (in parts per million i.e ppm) the data was collected for localities for 30 localities in a certain city and is presented below:

_{2}in the air (in parts per million i.e ppm) the data was collected for localities for 30 localities in a certain city and is presented below:

Concentration of SO_{2} (in ppm) | Frequency |

0.00 – 0.04 | 4 |

0.04 – 0.08 | 9 |

0.08 – 0.12 | 9 |

0.12 – 0.16 | 2 |

0.16 – 0.20 | 4 |

0.20 – 0.24 | 2 |

**Find the mean concentration of SO**_{2} in the air

_{2}in the air

**Solution:**

We may find class marks for each interval by using the relation

x = upperlimit + lowerclasslimit2x =

Class size of this data = 0.04

Now taking 0.04 assumed mean (x

_{i}) we may calculate d_{i}, u_{i}, f_{i}u_{i}as follows

Concentration of SO_{2}Frequency f_{1}Class interval x_{i`}d_{i}= x_{i}– 0.14u_{i}f_{i}u_{i}0.00 – 0.04 4 0.02 -0.12 -3 -12 0.04 – 0.08 9 0.06 -0.08 -2 -18 0.08 – 0.12 9 0.10 -0.04 -1 -9 0.12 – 0.16 2 0.14 0 0 0 0.16 – 0.20 4 0.18 0.04 1 4 0.20 – 0.24 2 0.22 0.08 2 4 TotalN = 30Sum = -31From the table we may observe that

N = 30

Sum = -31

= 0.14 + (0.04)(-31/30)

= 0.099 ppm

So mean concentration of SO

_{2}in the air is 0.099 ppm.

**Question 24. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.**

Number of days: | 0 – 6 | 6 – 10 | 10 – 14 | 14 – 20 | 20 – 28 | 28 – 38 | 38 – 40 |

Number of students: | 11 | 10 | 7 | 4 | 4 | 3 | 1 |

**Solution:**

We may find class mark of each interval by using the relation

x = upperlimit + lowerclasslimit2x =

Now, taking 16 as assumed mean (a) we may

Calculate d

_{i}and f_{i}d_{i}as follows

Number of daysNumber of students f_{i}X_{i}d = x_{i}+ 10f_{i}d_{i}0 – 6 11 3 -13 -143 6 – 10 10 8 -8 -280 10 – 14 7 12 -4 -28 14 – 20 7 16 0 0 20 – 28 8 24 8 32 28 – 36 3 33 17 51 30 – 40 1 39 23 23 TotalN = 40Sum = -145Now we may observe that

N = 40

Sum = -145

= 16 + (-145/40)

= 16 – 3.625

= 12.38

So mean number of days is 12.38 days, for which student was absent.

**Question 25. The following table gives the literacy rate (in percentage) of 35 cities. find the mean literacy rate.**

Literacy rate (in %): | 45 – 55 | 55 – 65 | 65 – 75 | 75 – 85 | 85 – 95 |

Number of cities: | 3 | 10 | 11 | 8 | 3 |

**Solution:**

We may find class marks by using the relation

x = upperlimit + lowerclasslimit2x =

Class size (h) for this data = 10

Now taking 70 as assumed mean (a) wrong

Calculate d

_{i}, u_{i}, f_{i}u_{i}as follows

Literacy rate (in %)Number of cities (f_{i})Mid value x_{i}d_{i}= x_{i}– 70u_{i}= d_{i}– 50f_{i}u_{i}45 – 55 3 50 -20 -20 -6 55 – 65 10 60 -10 -1 -10 65 – 75 11 70 0 0 0 75 – 85 8 80 10 1 8 85 – 95 3 90 20 2 6 TotalN = 35Sum = -2Now we may observe that

N = 35

Sum = -2

= 70 + (-2/35)

= 70 – 4/7

= 70 – 0.57

= 69.43

So, mean literacy rate is 69.43%

**Question 21. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contain varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.**

Number of mangoes: | 50 – 52 | 53 – 55 | 56 – 58 | 59 – 61 | 62 – 64 |

Number of boxes: | 15 | 110 | 135 | 115 | 25 |

**Find the mean number of mangoes kept in packing box. Which method of finding the mean did you choose?**

**Solution:**

Number of mangoesNumber of boxes50 – 52 15 53 – 55 110 56 – 58 135 59 – 61 115 62 – 64 25 We may observe that class internals are not continuous

There is a gap between two class intervals. So we have to add ½ from lower class limit to each interval and class mark (x

_{i}) may be obtained by using the relationx

_{i}= upperlimit + lowerclasslimit2Class size (h) of this data = 3

Now taking 57 as assumed mean (a) we may calculated d

_{i}, u_{i}, f_{i}u_{i}as follows

Class intervalFrequency f_{i}Mid values x_{i}d_{i}= x_{i}– A = x_{i}– 25f_{i}u_{i}49.5 – 52.5 15 51 -6 -2 -30 52.5 – 55.5 110 54 -3 -1 -110 55.5 – 58.5 135 57 0 0 0 58.5 – 61.5 115 60 3 1 115 61.5 – 64.5 25 63 6 2 50 TotalN = 400Sum = 25Now we have N

Sum = 25

Mean = A +h (sum/N)

= 57 + 3 (45/400)

= 57 + 3/16

= 57 + 0.1875

= 57.19

Clearly mean number of mangoes kept in packing box is 57.19

**Question 22. The table below shows the daily expenditure on food of 25 households in a locality**

Daily expenditure (In Rs): | 100 – 150 | 150 – 200 | 200 – 250 | 250 – 300 | 300 – 350 |

Number of households: | 4 | 5 | 12 | 2 | 2 |

**Find the mean daily expenditure on food by a suitable method.**

**Solution:**

We may calculate class mark (x

_{i}) for each interval by using the relationx

_{i}= upperlimit + lowerclasslimit2Class size = 50

Now, Taking 225 as assumed mean (x

_{i}) we may calculate d_{i}, u_{i}, f_{i}u_{i}as follows

Daily expenditureFrequency f_{1}Mid value x_{i}d_{i}= x_{i}– 225f_{i}u_{i}100 – 150 4 125 -100 -2 -8 150 – 200 5 175 -50 -1 -5 200 – 250 12 225 0 0 0 250 – 300 2 275 50 1 2 300 – 350 2 325 100 2 4 N = 25Sum = -7Now we may observe that

N = 25

Sum = -7

225 + 50 (-7/25)

225 – 14 = 211

So, mean daily expenditure on food is Rs 211

**Question 23. To find out the concentration of SO**_{2} in the air (in parts per million i.e ppm) the data was collected for localities for 30 localities in a certain city and is presented below:

_{2}in the air (in parts per million i.e ppm) the data was collected for localities for 30 localities in a certain city and is presented below:

Concentration of SO_{2} (in ppm) | Frequency |

0.00 – 0.04 | 4 |

0.04 – 0.08 | 9 |

0.08 – 0.12 | 9 |

0.12 – 0.16 | 2 |

0.16 – 0.20 | 4 |

0.20 – 0.24 | 2 |

**Find the mean concentration of SO**_{2} in the air

_{2}in the air

**Solution:**

We may find class marks for each interval by using the relation

x = upperlimit + lowerclasslimit2x =

Class size of this data = 0.04

Now taking 0.04 assumed mean (x

_{i}) we may calculate d_{i}, u_{i}, f_{i}u_{i}as follows

Concentration of SO_{2}Frequency f_{1}Class interval x_{i`}d_{i}= x_{i}– 0.14u_{i}f_{i}u_{i}0.00 – 0.04 4 0.02 -0.12 -3 -12 0.04 – 0.08 9 0.06 -0.08 -2 -18 0.08 – 0.12 9 0.10 -0.04 -1 -9 0.12 – 0.16 2 0.14 0 0 0 0.16 – 0.20 4 0.18 0.04 1 4 0.20 – 0.24 2 0.22 0.08 2 4 TotalN = 30Sum = -31From the table we may observe that

N = 30

Sum = -31

= 0.14 + (0.04)(-31/30)

= 0.099 ppm

So mean concentration of SO

_{2}in the air is 0.099 ppm.

**Question 24. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.**

Number of days: | 0 – 6 | 6 – 10 | 10 – 14 | 14 – 20 | 20 – 28 | 28 – 38 | 38 – 40 |

Number of students: | 11 | 10 | 7 | 4 | 4 | 3 | 1 |

**Solution:**

We may find class mark of each interval by using the relation

x = upperlimit + lowerclasslimit2x =

Now, taking 16 as assumed mean (a) we may

Calculate d

_{i}and f_{i}d_{i}as follows

Number of daysNumber of students f_{i}X_{i}d = x_{i}+ 10f_{i}d_{i}0 – 6 11 3 -13 -143 6 – 10 10 8 -8 -280 10 – 14 7 12 -4 -28 14 – 20 7 16 0 0 20 – 28 8 24 8 32 28 – 36 3 33 17 51 30 – 40 1 39 23 23 TotalN = 40Sum = -145Now we may observe that

N = 40

Sum = -145

= 16 + (-145/40)

= 16 – 3.625

= 12.38

So mean number of days is 12.38 days, for which student was absent.

**Question 25. The following table gives the literacy rate (in percentage) of 35 cities. find the mean literacy rate.**

Literacy rate (in %): | 45 – 55 | 55 – 65 | 65 – 75 | 75 – 85 | 85 – 95 |

Number of cities: | 3 | 10 | 11 | 8 | 3 |

**Solution:**

We may find class marks by using the relation

x = upperlimit + lowerclasslimit2x =

Class size (h) for this data = 10

Now taking 70 as assumed mean (a) wrong

Calculate d

_{i}, u_{i}, f_{i}u_{i}as follows

Literacy rate (in %)Number of cities (f_{i})Mid value x_{i}d_{i}= x_{i}– 70u_{i}= d_{i}– 50f_{i}u_{i}45 – 55 3 50 -20 -20 -6 55 – 65 10 60 -10 -1 -10 65 – 75 11 70 0 0 0 75 – 85 8 80 10 1 8 85 – 95 3 90 20 2 6 TotalN = 35Sum = -2Now we may observe that

N = 35

Sum = -2

= 70 + (-2/35)

= 70 – 4/7

= 70 – 0.57

= 69.43

So, mean literacy rate is 69.43%

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