# Class 10 RD Sharma Solutions – Chapter 7 Statistics – Exercise 7.1 | Set 2

**Problem 11: Five coins were simultaneously tossed 1000 times and at each toss, the number of heads was observed. The number of tosses during which 0, 1, 2, 3, 4**,** and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss.**

No. of heads per toss | No. of tosses |

0 | 38 |

1 | 144 |

2 | 342 |

3 | 287 |

4 | 164 |

5 | 25 |

Total | 1000 |

**Solution: **

No. of heads per toss (x) | No. of tosses (f) | fx |

0 | 38 | 0 |

1 | 144 | 144 |

2 | 342 | 684 |

3 | 287 | 861 |

4 | 164 | 656 |

5 | 25 | 125 |

Total (N) = 1000 | ∑ fx = 2470 |

We know that, Mean = ∑fx/ N = 2470/1000 = 2.47

So, the mean number of heads per toss are 2.47

**Problem 12: Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 50.**

x: | 10 | 30 | 50 | 70 | 90 | |

f: | 17 | f_{1} | 32 | f_{2} | 19 | Total = 120 |

**Solution:**

x | f | fx |

10 | 17 | 170 |

30 | f_{1} | 30f_{1} |

50 | 32 | 1600 |

70 | f_{2} | 70f_{2} |

90 | 19 | 1710 |

N = 68 + f_{1} + f2 = 120 | ∑ fx = 30f_{1} + 70f_{2 }+ 3480 _{ } |

Given,

Mean = 50, N = 120

We know that,

Mean = ∑fx/ N = (30f

_{1}+ 70f_{2}+ 3480)/(68 + f_{1}+ f_{2})Now,

50 = (30f

_{1}+ 70f_{2}+ 3480)/(68 + f_{1}+ f_{2})Also , 68 + f

_{1}+ f_{2}= 120f

_{1}= 52 – f_{2}……. (i)Now,

50 = (30f

_{1}+ 70f_{2}+ 3480)/ 12030f

_{1}+ 70f_{2 }= 6000 – 3480Now, putting the value of f

_{1}from equation (i) –30(52 – f

_{2}) + 70f_{2 }= 25201560 – 30f

_{2 }+ 70f_{2}= 252040f

_{2}= 960So, f

_{2}= 24and f

_{1}= 52 – f_{2}= 52 – 24 = 28Thus, f

_{1}= 28 and f_{2}= 24

**Problem 13: The arithmetic mean of the following data is 14, find the value of k.**

x_{i }: | 5 | 10 | 15 | 20 | 15 |

f_{i }: | 7 | k | 8 | 4 | 5 |

**Solution:**

x | f | fx |

5 | 7 | 35 |

10 | k | 10k |

15 | 8 | 120 |

20 | 4 | 80 |

25 | 5 | 125 |

N = k + 24 | ∑ fx = 360 + 10k |

Given,

Mean = 14

We know that,

Mean = ∑fx/ N = (360 + 10k)/(k + 24)

Now,

14(k + 24) = 360 + 10k

14k + 336 = 360 + 10k

4k = 24

So, k = 6

**Problem 14: The arithmetic mean of the following data is 25, find the value of k.**

x_{i} : | 5 | 15 | 25 | 35 | 45 |

f_{i} : | 3 | k | 3 | 6 | 2 |

**Solution: **

x | f | fx |

5 | 3 | 15 |

15 | k | 15k |

25 | 3 | 75 |

35 | 6 | 210 |

45 | 2 | 90 |

N = 14 + k | ∑ fx = 390 + 15k |

Given,

Mean = 25

We know that,

Mean = ∑fx/ N = (390 + 15k)/(k + 14)

Now,

25(k + 14) = 390 + 15k

25k + 350 = 390 + 15k

10k = 40

So, k = 4

**Problem 15: If the mean of the following data is 18.75. Find the value of p.**

x_{i} : | 10 | 15 | p | 25 | 30 |

f_{i} : | 5 | 10 | 7 | 8 | 2 |

**Solution: **

x | f | fx |

10 | 5 | 50 |

15 | 10 | 150 |

p | 7 | 7p |

25 | 8 | 200 |

30 | 2 | 60 |

N = 32 | ∑ fx = 460 + 7p |

Given,

Mean = 18.75

We know that,

Mean = ∑fx/ N = (460 + 7p)/ 32

Now,

18.75 × 32 = 460 + 7p

600 = 460 + 7p

7p = 140

So, p = 20

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