# RS Aggarwal Class 9 Solutions Chapter 14 -Statistics Ex 14E (14.5)

## RS Aggarwal Class 9 Ex 14E Chapter 14

Q.1: Find the mean of daily wages of 60 workers in a factory as per data given below:

 Daily wage ( in Rs.) (xi$x_{i}$) 90 110 120 130 150 No. of workers (fi$f_{i}$) 12 14 13 11 10

Sol:

For calculating the mean, we prepare the following table:

 Daily wage (in Rs.) (xi$x_{i}$) No. of workers (fi$f_{i}$) fixi$f_{i}x_{i}$ 90 12 1080 110 14 1540 120 13 1560 130 11 1430 150 10 1500 Σfi=60$\Sigma f_{i}= 60$ Σfixi$\Sigma f_{i}x_{i}$=7110

Mean = ΣfixiΣfi=711060=118.5$\frac{\Sigma f_{i}x_{i}}{\Sigma f_{i}}=\frac{7110}{60}= 118.5$

Therefore, the mean of daily wages of 60 workers = Rs. 118.50

Q.2: The following table shows the weight of 12 workers in a factory:

 Weight ( in kg) (xi$x_{i}$) 60 63 66 69 72 No. of workers (fi$f_{i}$) 4 3 2 2 1

Find the mean weight.

Sol:

For calculating the mean, we prepare the following frequency table:

 Weight ( in kg) (xi$x_{i}$) No. of workers (fi$f_{i}$) fixi$f_{i}x_{i}$ 60 4 240 63 3 189 66 2 132 69 2 138 72 1 72 Σfi=12$\Sigma f_{i}= 12$ Σfixi$\Sigma f_{i}x_{i}$=771

Mean = ΣfixiΣfi=77112=62.25$\frac{\Sigma f_{i}x_{i}}{\Sigma f_{i}}=\frac{771}{12}= 62.25$ kg

Therefore, mean of workers = 62.25 kg

Q.3: The following data gives the number of boys of a particular age in a class of 40 students.

 Age ( in years ) (xi$x_{i}$) 15 16 17 18 19 20 Frequency (fi$f_{i}$) 3 8 9 11 6 3

Calculate the mean age of the students.

Sol:

For calculating the mean, we prepare the following frequency table:

 Age ( in years ) (xi$x_{i}$) Frequency (fi$f_{i}$) fixi$f_{i}x_{i}$ 15 3 45 16 8 128 17 9 153 18 11 198 19 6 114 20 3 60 Σfi=12$\Sigma f_{i}= 12$ Σfixi$\Sigma f_{i}x_{i}$=698

Mean = ΣfixiΣfi=69840=17.45$\frac{\Sigma f_{i}x_{i}}{\Sigma f_{i}}=\frac{698}{40}= 17.45$ kg

Therefore, the mean age of the students = 17.45 kg

Q.4: Find the mean of the following frequency distribution:

 Variable (xi$x_{i}$) 10 30 50 70 89 Frequency (fi$f_{i}$) 7 8 10 15 10

Sol:

For calculating the mean, we prepare the following frequency table:

 Variable (xi$x_{i}$) Frequency (fi$f_{i}$) fixi$f_{i}x_{i}$ 10 7 70 30 8 240 50 10 500 70 15 1050 89 10 890 Σfi=50$\Sigma f_{i}= 50$ Σfixi$\Sigma f_{i}x_{i}$=2750

Mean = ΣfixiΣfi=275050=55$\frac{\Sigma f_{i}x_{i}}{\Sigma f_{i}}=\frac{2750}{50}= 55$.

Q.5: If the mean of the following frequency distribution is 8, find the value of p.

 (xi$x_{i}$) 3 5 7 9 11 13 (fi$f_{i}$) 6 8 15 p 8 4

We prepare the following frequency table:

 (xi$x_{i}$) (fi$f_{i}$) fixi$f_{i}x_{i}$ 3 6 18 5 8 40 7 15 105 9 p 9p 11 8 88 13 4 52 Σfi=41+p$\Sigma f_{i}= 41 + p$ Σfixi$\Sigma f_{i}x_{i}$= 303+9p

Mean = ΣfixiΣfi=303+9p41+p$\frac{\Sigma f_{i}x_{i}}{\Sigma f_{i}}=\frac{303+9p}{41+p}$

But, mean = 8 (given)

Therefore, 303+9p41+p=8$\frac{303+9p}{41+p} =8$

303+9p=8(41+p)$\Rightarrow 303+9p =8(41+p)$
303+9p=328+8p$\Rightarrow 303+9p = 328+8p$

9p8p=328303$\Rightarrow 9p -8p = 328-303$ = 25

Therefore, the value of p = 25

Q.6: Find the missing frequency p for the following frequency distribution whose mean is 28.25.

 (xi$x_{i}$) 15 20 25 30 35 40 (fi$f_{i}$) 8 7 p 14 15 16

Sol:

We prepare the frequency distribution table:

 (xi$x_{i}$) (fi$f_{i}$) fixi$f_{i}x_{i}$ 15 6 18 20 8 40 25 15 105 30 p 9p 35 8 88 40 4 52 13 Σfi=50+p$\Sigma f_{i}= 50 + p$ Σfixi$\Sigma f_{i}x_{i}$= 1445+25p

Mean = ΣfixiΣfi=1445+25p50+p$\frac{\Sigma f_{i}x_{i}}{\Sigma f_{i}}=\frac{1445+25p}{50+p}$

But, mean = 28.25 (given)

Therefore, 1445+25p50+p=8$\frac{1445+25p}{50+p} =8$

1445+25p=(50+p)(28.25)$\Rightarrow 1445 + 25p =(50+p)(28.25)$
1445+25p=1412.50+28.25p$\Rightarrow 1445 + 25p = 1412.50 + 28.25p$
28.25p+25p=1445+1412.50$\Rightarrow -28.25p +25p = -1445 +1412.50$

3.25p=32.5$\Rightarrow -3.25p = -32.5$ = 10

Therefore, the value of p = 10

Q.7: Find the value of p for the following frequency distribution whose mean is 16.6 .

 (xi$x_{i}$) 8 12 15 p 20 25 40 (fi$f_{i}$) 12 16 20 24 16 8 4

Sol:

We prepare the following frequency distribution table:

 (xi$x_{i}$) (fi$f_{i}$) fixi$f_{i}x_{i}$ 8 12 96 12 16 192 15 20 300 p 24 24p 20 16 320 25 8 200 30 4 120 Σfi=100$\Sigma f_{i}= 100$ Σfixi$\Sigma f_{i}x_{i}$= 1228+24p

Mean = ΣfixiΣfi=1228+24p100$\frac{\Sigma f_{i}x_{i}}{\Sigma f_{i}}=\frac{1228+24p}{100}$

But mean = 16.6 (given)

Therefore, 1228+24p100=16.6$\frac{1228+24p}{100} =16.6$

1228+24p=1660$\Rightarrow 1228 + 24p =1660$
24p=432$\Rightarrow 24p = 432$

p=43224$\Rightarrow p=\frac{432}{24}$ =18

Therefore, the value of p = 18

Q.8: Find the missing frequencies in the following frequency distribution, whose mean is 50.

 xi$x_{i}$ 10 30 50 70 90 Total fi$f_{i}$ 17 f1$f_{1}$ 32 f2$f_{2}$ 19 120

Sol:

Let, f1$f_{1}$ and f2$f_{2}$ be the missing frequencies.

We prepare the following frequency distribution table:

 xi$x_{i}$ fi$f_{i}$ fixi$f_{i}x_{i}$ 10 17 170 30 f1$f_{1}$ 30f1$30f_{1}$ 50 32 1600 70 f2$f_{2}$ 70f2$70f_{2}$ 90 19 1710 Total 120 3480+30f1$30f_{1}$+70f2$70f_{2}$

Here, Σfi=68+f1+f2$\Sigma f_{i}=68+f_{1}+f_{2}$

But, 68+f1+f2$68+ f_{1}+f_{2}$= 120    (given)

Therefore, f1+f2$f_{1}+f_{2}$= 120-68 = 52

Thus, f2=52f1$f_{2}=52-f_{1}$ . . . . . . . . . (i)

Also,

Mean = fixifi=3480+30f1+70f2120$\frac{f_{i}x_{i}}{f_{i}}=\frac{3480+30f_{1}+70f_{2}}{120}$

= 3480+30f1+70(52f1)120$\frac{3480+30f_{1}+70(52-f_{1})}{120}$   (using equation (i) )

= 3480+30f1+364070f1120$\frac{3480+30f_{1}+3640-70f_{1}}{120}$

= 712040f1120$\frac{7120-40f_{1}}{120}$

But mean = 50 (given)

Thus we have:

50=712040f1120$50=\frac{7120-40f_{1}}{120}$
6000=712040f1$6000=7120-40f_{1}$
40f1=71206000$40f_{1}=7120-6000$
40f1=1120$40f_{1}=1120$
f1=112040=28$f_{1}=\frac{1120}{40}=28$

Substituting the value of f1$f_{1}$ in equation 1, we have,

f2$f_{2}$= 52 – 28 = 24

Thus, the missing frequencies are f1$f_{1}$=28 and f2$f_{2}$= 24.

Q.9: Use the assumed- mean method to find the mean weekly wages from the data given below:

 Weekly wages xi$x_{i}$ 800 820 860 900 920 980 1000 No. of workers fi$f_{i}$ 7 14 19 25 20 10 5

Sol:

Let, us assume the mean (A) = 900

 Weekly wages xi$x_{i}$ No. of workers fi$f_{i}$ di=xi−A$d_{i}=x_{i}-A$ =xi−900$x_{i}-900$ fi×di$f_{i}\times d_{i}$ 800 7 -100 -700 820 14 -80 -1120 860 19 -40 -760 900 25 0 0 920 20 20 400 980 10 80 800 1000 5 100 500 Σfi=100$\Sigma f_{i}= 100$ -880

Let, X¯$\bar{X}$ be the mean.

Using formula: X¯=A+Σfi×diΣfi$\bar{X}=A+\frac{\Sigma f_{i}\times d_{i}}{\Sigma f_{i}}$

= 900+880100$900+\frac{-880}{100}$ = 900-8.80 = 891.20

Therefore, the mean weekly wages = Rs. 891.20

Q.10: Use the assumed-mean method to find the mean height of the plants from the following frequency-distribution table.

 Height (in cm) x{i} 61 64 67 70 73 No. of plants f{i} 5 18 42 27 8

Sol:

Let, the assumed mean be A= 67

 Height (in cm) x{i} No. of plants f_{i} di=xi−A$d_{i}=x_{i}-A$ =xi−67$x_{i}-67$ fi×di$f_{i}\times d_{i}$ 61 5 -6 -30 64 18 -3 -54 67 42 0 0 70 27 3 81 73 8 6 48 100 Σfidi=45$\Sigma f_{i}d_{i}=45$

Let, X¯$\bar{X}$ be the mean.

Using formula: X¯=A+Σfi×diΣfi$\bar{X}=A+\frac{\Sigma f_{i}\times d_{i}}{\Sigma f_{i}}$

= 67+45100$67+\frac{45}{100}$ = 67+0.45 = 67.45

Therefore, the mean height of the plant = 67.45 cm.

Q11: Use the step-deviation method to find the arithmetic mean from the following data.

 xi$x_{i}$ 18 19 20 21 22 23 24 fi$f_{i}$ 170 320 530 700 230 140 110

Sol:

Clearly, h=1 . Let the asumed mean A=21

 xi$x_{i}$ fi$f_{i}$ ui=xi−211$u_{i}=\frac{x_{i}-21}{1}$ fiui$f_{i}u_{i}$ 18 170 -3 -510 19 320 -2 -640 20 530 -1 -530 21 700 0 0 22 230 1 230 23 140 2 280 24 110 3 330 Total Σfi=2200$\Sigma f_{i}=2200$ Σfiui=−840$\Sigma f_{i}u_{i}=-840$

Let, X¯$\bar{X}$ be the mean.

Using formula: X¯=A+h×ΣfiuiΣfi$\bar{X}=A+h\times \frac{\Sigma f_{i}u_{i}}{\Sigma f_{i}}$

X¯=21+1×8402200$\bar{X}=21+1\times \frac{-840}{2200}$

X¯=210.38$\bar{X}=21-0.38$ = 20.62

Thus, the mean is 20.62

Q.12: The table below gives the distribution of villages and their heights from the sea level in a certain region.

 (Height in m) xi$x_{i}$ 200 600 1000 1400 1800 2200 No. of villages fi$f_{i}$ 142 265 560 271 89 16

Sol:

Clearly, h = (x2x1$x_{2}-x_{1}$) = 600 – 200 = 400

Let’s assume that mean A = 1000

 (Height in m) xi$x_{i}$ No. of villages fi$f_{i}$ ui=xi−211$u_{i}=\frac{x_{i}-21}{1}$ fiui$f_{i}u_{i}$ 200 142 -2 -284 600 265 -1 -265 1000 560 0 0 1400 271 1 271 1800 89 2 178 2200 16 3 48 Total Σfi=1343$\Sigma f_{i}=1343$ Σfiui=−52$\Sigma f_{i}u_{i}=-52$

Let, X¯$\bar{X}$ be the mean.

Using formula: X¯=A+h×ΣfiuiΣfi$\bar{X}=A+h\times \frac{\Sigma f_{i}u_{i}}{\Sigma f_{i}}$

X¯=1000+400×521343$\bar{X}=1000+400 \times \frac{-52}{1343}$

X¯=100015.488$\bar{X}=1000-15.488$ = 984.51

Thus, the mean height is 984.51m.

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