RS Aggarwal Solutions Class 9 Ex 14E

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)

90 110 120 130 150
No. of workers

(fi)

12 14 13 11 10

Sol:

For calculating the mean, we prepare the following table:

Daily wage (in Rs.)

(xi)

No. of workers

(fi)

fixi
90 12 1080
110 14 1540
120 13 1560
130 11 1430
150 10 1500
Σfi=60 Σfixi=7110

Mean = ΣfixiΣfi=711060=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)

60 63 66 69 72
No. of workers

(fi)

4 3 2 2 1

Find the mean weight.

Sol:

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

Weight ( in kg)

(xi)

No. of workers

(fi)

fixi
60 4 240
63 3 189
66 2 132
69 2 138
72 1 72
Σfi=12 Σfixi=771

Mean = ΣfixiΣfi=77112=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)

15 16 17 18 19 20
Frequency

(fi)

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)

Frequency

(fi)

fixi
15 3 45
16 8 128
17 9 153
18 11 198
19 6 114
20 3 60
Σfi=12 Σfixi=698

Mean = ΣfixiΣfi=69840=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)

10 30 50 70 89
Frequency

(fi)

7 8 10 15 10

Sol:

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

Variable

(xi)

Frequency

(fi)

fixi
10 7 70
30 8 240
50 10 500
70 15 1050
89 10 890
Σfi=50 Σfixi=2750

Mean = ΣfixiΣfi=275050=55.

 

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

(xi) 3 5 7 9 11 13
(fi) 6 8 15 p 8 4

We prepare the following frequency table:

(xi) (fi) fixi
3 6 18
5 8 40
7 15 105
9 p 9p
11 8 88
13 4 52
Σfi=41+p Σfixi= 303+9p

Mean = ΣfixiΣfi=303+9p41+p

But, mean = 8 (given)

Therefore, 303+9p41+p=8

303+9p=8(41+p)
303+9p=328+8p

9p8p=328303 = 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) 15 20 25 30 35 40
(fi) 8 7 p 14 15 16

Sol:

We prepare the frequency distribution table:

(xi) (fi) fixi
15 6 18
20 8 40
25 15 105
30 p 9p
35 8 88
40 4 52
13 Σfi=50+p Σfixi= 1445+25p

Mean = ΣfixiΣfi=1445+25p50+p

But, mean = 28.25 (given)

Therefore, 1445+25p50+p=8

1445+25p=(50+p)(28.25)
1445+25p=1412.50+28.25p
28.25p+25p=1445+1412.50

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) 8 12 15 p 20 25 40
(fi) 12 16 20 24 16 8 4

Sol:

We prepare the following frequency distribution table:

(xi) (fi) fixi
8 12 96
12 16 192
15 20 300
p 24 24p
20 16 320
25 8 200
30 4 120
Σfi=100 Σfixi= 1228+24p

Mean = ΣfixiΣfi=1228+24p100

But mean = 16.6 (given)

Therefore, 1228+24p100=16.6

1228+24p=1660
24p=432

p=43224 =18

Therefore, the value of p = 18

 

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

xi 10 30 50 70 90 Total
fi 17 f1 32 f2 19 120

Sol:

Let, f1 and f2 be the missing frequencies.

We prepare the following frequency distribution table:

xi fi fixi
10 17 170
30 f1 30f1
50 32 1600
70 f2 70f2
90 19 1710
Total 120 3480+30f1+70f2

Here, Σfi=68+f1+f2

But, 68+f1+f2= 120    (given)

Therefore, f1+f2= 120-68 = 52

Thus, f2=52f1 . . . . . . . . . (i)

Also,

Mean = fixifi=3480+30f1+70f2120

= 3480+30f1+70(52f1)120   (using equation (i) )

= 3480+30f1+364070f1120

= 712040f1120

But mean = 50 (given)

Thus we have:

50=712040f1120
6000=712040f1
40f1=71206000
40f1=1120
f1=112040=28

Substituting the value of f1 in equation 1, we have,

f2= 52 – 28 = 24

Thus, the missing frequencies are f1=28 and f2= 24.

 

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

Weekly wages

xi

800 820 860 900 920 980 1000
No. of workers

fi

7 14 19 25 20 10 5

Sol:

Let, us assume the mean (A) = 900

Weekly wages

xi

No. of workers

fi

di=xiA

=xi900

fi×di
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 -880

Let, X¯ be the mean.

Using formula: X¯=A+Σfi×diΣfi

= 900+880100 = 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=xiA

=xi67

fi×di
61 5 -6 -30
64 18 -3 -54
67 42 0 0
70 27 3 81
73 8 6 48
100 Σfidi=45

Let, X¯ be the mean.

Using formula: X¯=A+Σfi×diΣfi

= 67+45100 = 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 18 19 20 21 22 23 24
fi 170 320 530 700 230 140 110

Sol:

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

xi fi ui=xi211 fiui
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 Σfiui=840

Let, X¯ be the mean.

Using formula: X¯=A+h×ΣfiuiΣfi

X¯=21+1×8402200

X¯=210.38 = 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 200 600 1000 1400 1800 2200
No. of villages

fi

142 265 560 271 89 16

Sol:

Clearly, h = (x2x1) = 600 – 200 = 400

Let’s assume that mean A = 1000

(Height in m) xi No. of villages

fi

ui=xi211 fiui
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 Σfiui=52

Let, X¯ be the mean.

Using formula: X¯=A+h×ΣfiuiΣfi

X¯=1000+400×521343

X¯=100015.488 = 984.51

Thus, the mean height is 984.51m.


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