# RD Sharma Solutions Class 9 Measures Of Central Tendency Exercise 24.2

## RD Sharma Solutions Class 9 Chapter 24 Exercise 24.2

### RD Sharma Class 9 Solutions Chapter 24 Ex 24.2 Free Download

Q 1 . Calculate the mean for the following distribution :

x :           5              6              7              8              9

f :            4              8              14           11           3

SOLUTION :

 x f fx 5 6 7 8 9 4 8 14 11 3 20 48 98 88 27 N=40 $\sum fx=281$

$∴ Mean\;\bar{x}=\frac{\sum fx }{N}$

=$\frac{281}{40}$ = 7.025.

Q 2 . Find the mean of the following data :

x :           19           21           23           25           27           29           31

f :            13           15           16           18           16           15           13

SOLUTION :

 x f fx 19 21 23 25 27 29 31 13 15 16 18 16 15 13 247 315 368 450 432 435 403 N=106 $\sum fx=2650$

$∴ Mean\;\bar{x}=\frac{\sum fx }{N}$

=$\frac{2650}{106}$ = 25.

Q 3 . The mean of the following data is 20.6 .Find the value of p.

x :           10           15           p             25           35

f :            3              10           25           7              5

SOLUTION :

 x f fx 10 15 P 25 35 3 10 25 7 5 30 150 25p 175 175 N = 50 $\sum fx=25p+530$

It is given that ,

Mean = 20.6

$\Rightarrow \frac{\sum fx}{N}=20.6$

$\Rightarrow \frac{25p+530}{50}=20.6$

$\Rightarrow 25p+530=20.6\times 50$

$\Rightarrow 25p=1030-530$

$\Rightarrow 25p=500$

$\Rightarrow p=\frac{500}{25}$ =20

$\Rightarrow p=20$

$∴ p=20$.

Q 4 . If the mean of the following data is 15 , find p.

x :           5              10           15           20           25

f :            6              p             6              10           5

SOLUTION :

 x f fx 5 10 15 20 25 6 P 6 10 5 30 10p 90 200 125 N=p+27 $\sum fx=10p+445$

It is given that ,

Mean = 15

$\Rightarrow \frac{\sum fx}{N}=15$

$\Rightarrow \frac{10p+445}{p+27}=15$

$\Rightarrow 10p+445=15\times \left ( p+27 \right )$

$\Rightarrow 10p +445=15p+405$

$\Rightarrow 15p-10p=445-405$

$\Rightarrow 5p=40$

$\Rightarrow p=\frac{40}{5}$ =8

$\Rightarrow p=8$

$∴ p=8$.

Q 5 . Find the value of p for the following distribution whose mean is 16.6.

x :           8              12           15           p             20           25           30

f :            12           16           20           24           16           8              4

SOLUTION :

 x f fx 8 12 15 P 20 25 30 12 16 20 24 16 8 4 96 192 300 24p 320 200 120 N=100 $\sum fx=24p+1228$

It is given that ,

Mean = 16.6

$\Rightarrow \frac{\sum fx}{N}=16.6$

$\Rightarrow \frac{24p+1228}{100}=16.6$

$\Rightarrow 24p+1228=1660$

$\Rightarrow 24p =1660-1228$

$\Rightarrow 24p=432$

$\Rightarrow p=\frac{432}{24}$ =18

$\Rightarrow p=18$

$∴ p=18$.

Q 6 . Find the missing value of p for the following distribution whose mean is 12.58 .

x :           5              8              10           12           p             20           25

f :            2              5              8              22           7              4              2

SOLUTION :

 x f fx 5 8 10 12 P 20 25 2 5 8 22 7 4 2 10 40 80 264 7p 80 50 N = 50 $\sum fx=7p+524$

It is given that ,

Mean = 12.58

$\Rightarrow \frac{\sum fx}{N}=12.58$

$\Rightarrow \frac{7p+524}{50}=12.58$

$\Rightarrow 7p+524=629$

$\Rightarrow 7p =629-524$

$\Rightarrow 7p=105$

$\Rightarrow p=\frac{105}{7}$ =15

$\Rightarrow p=15$

$∴ p=18$.

Q 7 . Find the missing frequency (p) for the following distribution whose mean is 7.68 .

x :           3              5              7              9              11           13

f :            6              8              15           p             8              4

SOLUTION :

 x f fx 3 5 7 9 11 13 6 8 15 P 8 4 18 40 105 9p 88 52 N=p+41 $\sum fx=9p+303$

It is given that ,

Mean = 7.68

$\Rightarrow \frac{\sum fx}{N}=7.68$

$\Rightarrow \frac{9p+303}{ p+41}=7.68$

$\Rightarrow 9p+303=7.68p +314.88$

$\Rightarrow 9p-7.68p =314.88-303$

$\Rightarrow 1.32p=11.88$

$\Rightarrow p=\frac{11.88}{1.32}$ =9

$\Rightarrow p=9$

$∴ p=9$.

Q 8 . Find the value of p , if the mean of the following distribution is 20 .

x :           15           17           19           20+p      23

f :            2              3              4              5p           6

SOLUTION :

 x f fx 15 17 19 20+p 23 2 3 4 5p 6 30 51 76 100p+ $5p^{2}$ 138 N=5p+15 fx= $5p^{2}+100p+295$

It is given that ,

Mean = 20

$\Rightarrow \frac{\sum fx}{N}=20$

$\Rightarrow\frac{ 5p^{2}+100p+295}{5p+15}=20$

$\Rightarrow5p^{2}+100p+295=20\left ( 5p+15 \right )$

$\Rightarrow5p^{2}+100p+295=100p+300$

$\Rightarrow5p^{2}=300-295$

$\Rightarrow5p^{2}=5$

$\Rightarrow p^{2}=1$

$\Rightarrow p=\pm 1$

Frequency can’t be negative.

Hence, value of p is 1.

Q 9 . Find the mean of the following distribution :

x :           10           12           20           25           35

f :            3              10           15           7              5

SOLUTION :

 x f fx 10 12 20 25 35 3 10 15 7 5 30 120 300 175 175 N=40 $\sum fx=800$

$∴ Mean\;\bar{x}=\frac{\sum fx }{N}$

=$\frac{800}{40}$ = 20.

Q 10. Candidates of four schools appear in a mathematics test. The data were as follows :

 Schools No. Of Candidates Average Score I II III   IV 60 48 Not Available 40 75 80 55   50

If the average score of the candidates of all four schools is 66 , Find the number of candidates that appeared from school  III .

SOLUTION :

 Schools No. Of Candidates Average Score I II III IV 60 48 x 40 75 80 55 50

Given the average score of all schools =66

$\Rightarrow \frac{N_{1}\bar{x}_{1}+N_{2}\bar{x}_{2}+N_{3}\bar{x}_{3}+N_{4}\bar{x}_{4}}{N_{1}+N_{2}+N_{3}+N_{4}}=66$

$\Rightarrow \frac{60\times 75+48\times 80+x\times 55+40\times 50}{60+48+x+40}=66$

$\Rightarrow \frac{4500+3840+55x+2000}{148+x}=66$

$\Rightarrow \frac{10340+55x}{148+x}=66$

$\Rightarrow 10340+55x=66x+9768$

$\Rightarrow 10340-9768=66x-55x$

$\Rightarrow 11x=572$

$\Rightarrow x=\frac{572}{11}=52$

$∴$ No. of candidates appeared from school III = 52.

Q 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 1 2 3 4 5 38 144 342 287 164 25 Total 1000

SOLUTION :

 No . of heads per toss(x) No.of tosses(f) fx 0 1 2 3 4 5 38 144 342 287 164 25 0 144 684 861 656 125 N=1000 $\sum fx=2470$

$∴ Mean$ number of heads per toss = $\frac{\sum fx}{N}$

= $\frac{2470}{1000}$

= 2.47

Q 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 30 50 70 90 17 $f_{1}$ 32 $f_{2}$ 19 170 30$f_{1}$ 1600 70$f_{2}$ 1710 N=120 $\sum fx=3480+30f_{1}+70f_{2}$

It is given that

Mean = 50

$\Rightarrow \frac{\sum fx}{N}=50$

$\Rightarrow \frac{3480+30f_{1}+70f_{2}}{N}$=50

$\Rightarrow 3480+30f_{1}+70f_{2}=50\times 120$

$\Rightarrow 30f_{1}+70f_{2}=6000-3480$

$\Rightarrow 10\left (3f_{1}+7f_{2} \right )=10\left (252 \right )$

$\Rightarrow 3f_{1}+7f_{2} =252 \cdot \cdot \cdot \cdot \cdot \left ( 1 \right )$                         $\left [∵ Divide \;by \;10 \right ]$

And N = 20

$\Rightarrow 17+f_{1}+32+f_{2}+19=120$

$\Rightarrow 68+f_{1}+f_{2}=120$

$\Rightarrow f_{1}+f_{2}=120-68$

$\Rightarrow f_{1}+f_{2}=52$

Multiply with 3 on both sides

$\Rightarrow 3f_{1}+3f_{2}=156\cdot \cdot \cdot \cdot \cdot \cdot \left ( 2 \right )$

Subtracting equation (2) from equation (1)

$\Rightarrow 3f_{1}+7f_{2}-3f_{1}-3f_{2}=252-156$

$\Rightarrow 4f_{2}=96$

$\Rightarrow f_{2}=\frac{96}{4}$ = 24

Put the value of $f_{2}$ in equation (1)

$\Rightarrow 3f_{1}+7\times 24=252$

$\Rightarrow 3f_{1}=252-168$

$\Rightarrow f_{1}=\frac{84}{3}=28$

$\Rightarrow f_{1}=28$