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Chapter 32: Statistics – Exercise 32.4

Statistics – Exercise – 32.4 – Q.1(i)

x d = (x - Mean) d2
2 -5 25
4 -3 9
5 -2 4
6 -1 1
8 1 1
17 10 100
42   140

Statistics – Exercise – 32.4 – Q.1(i)

 

Statistics – Exercise – 32.4 – Q.1(ii)

x d = (x - Mean) d2
6 -3 9
7 -2 4
10 1 1
12 3 9
13 4 16
4 -5 25
8 -1 1
12 3 9
72   74

Statistics – Exercise – 32.4 – Q.1(ii)

 

Statistics – Exercise – 32.4 – Q.1(iii)

xi di = xi - 299 d12
227 -72 5184
235 -64 4096
255 -44 1936
269 -30 900
292 -7 49
299 0 0
312 13 169
321 22 484
333 34 1156
348 49 2401
  Total = – 99 Total = 16375

Statistics – Exercise – 32.4 – Q.1(iii)

 

Statistics – Exercise – 32.4 – Q.1(iv)

xi di = x- 299      d12
15 0 0
22 7 49
27 12 144
11 -4 16
9 -6 36
21 6 36
14 -1 1
9 -6 36
     
  Total = – 8 Total = 318

Statistics – Exercise – 32.4 – Q.1(iv)

 

Statistics – Exercise – 32.4 – Q.2

We have, n = 20, and σ2 = 5

Now each observation is multiplied by 2.

Suppose X = 2x be the new data.

Statistics – Exercise – 32.4 – Q.2

Since, σ2 = 5

Now, for the new data:

 

Statistics – Exercise – 32.4 – Q.3

We have, n = 15, and σ2 = 4

Now each observation is increased by 9.

Suppose x = x + 9 be the new data.

Since, σ2 = 5

Now, for the new data:

 

Statistics – Exercise – 32.4 – Q.4

Let the other two be x and y

1 + 2 + 6 + x + y = 5 * 4.4 because of the mean

x + y = 13

Variance = [(1 - 4.4)2 + (2 - 4.4)2 + (6 - 4.4)2 + (x - 4.4)2 + (y - 4.4)2]/5

Hence

11.56 + 5.76 + 2.56 + (x - 4.4)2 + (y - 4.4)2 = 41.2

(x - 4.4)2 + (13 - x - 4.4)2 = 21.32

x2 - 8.8x + 19.36 + 73.96 - 17.2x + x2 = 21.32

2x2 - 26x + 72 = 0

x2 - 13x + 36 = 0

(x - 4)(x - 9) = 0

x = 4 or x = 9

If x = 4, y = 9 and

The other two observations are 4 and 9.

 

Statistics – Exercise – 32.4 – Q.5

If mean and SD of observations areand σ respectively, then mean and SD of observations multiplied by a constant 'k' are

SD = |k|σ

In this question, it is given that k = 3

So New mean = 8 × 3 = 24

New SD = 4 × 3 = 12 

 

Statistics – Exercise – 32.4 – Q.6

Let x and y be the remaining two observations. Then,

Mean = 9

⟹ 60 + x + y + = 72

⟹ x + y = 12 ..... (i)

Variance = 9.25

⟹ 642 + x2 + y2 = 722

⟹ x2 + y2 = 80 ..... (ii)

Now, (x + y)2 + (x - y)2 = 2(x2 + y2)

⟹ 144 + (x - y)2 = 2 × 80

⟹ (x - y)2 = 16

⟹ x - y = ±4

if x - y  = 4, then x + y = 12 and x - y = 4 ⟹ x = 8, y = 4

if x - y = -4, then x + y = 12 and x - y = -4 ⟹ x = 4, y = 8

Hence, the remaining two observations are 4 and 8.

 

Statistics – Exercise – 32.4 – Q.7

We have,

n = 200, = 40, σ = 15

Corrected Σxi = Incorrect Σxi - (Sum of incorrect values) + (Sum of correct values)

= 8000 - 34 - 53 + 43 + 35 = 7991

Statistics – Exercise – 32.4 – Q.7(i)

= 365000 - (34)2 - 532 + (43)2 + 352 = 364109

891

 

Statistics – Exercise – 32.4 – Q.8

We have,

n = 100,= 40, σ = 5.1

Corrected Σxi = incorrect Σxi - (Sum of incorrect values) + (Sum of correct values)

= 4000 - 50 + 40 = 3990

Statistics – Exercise – 32.4 – Q.8(i)

= 162601 - (50)2 + (40)2 = 161701

 

Statistics – Exercise – 32.4 – Q.9

We have, 

n = 20,= 10 and σ = 2

⟹ Incorrected Σxi = 200

and,

Statistics – Exercise – 32.4 – Q.9(ii)

(i) When 8 is omitted from the data:

If 8 is omitted from the data, then 19 observations are left.

Now, Incorrected Σxi = 200

⟹ Corrected Σxi + 8 = 200

⟹ Corrected Σxi = 192

and, 

Statistics – Exercise – 32.4 – Q.9(iii)

(ii) When the incorrect observation 8 is replaced by 12:

we have, Incorrected Σxi = 200

∴ Corrected Σxi = 200 - 8 + 12 = 204

and,

Statistics – Exercise – 32.4 – Q.9(iv)

 

Statistics – Exercise – 32.4 – Q.10

We have, 

⟹ Incorrect Σxi = 2000

and,

Statistics – Exercise – 32.4 – Q.10

When the incorrect observations 21, 21, 18 are omitted from the data:

n = 97

Now, Incorrect Σxi = 2000

⟹ Corrected Σxi = 2000 - 21 - 21 - 18 = 1940

and, 

Statistics – Exercise – 32.4 – Q.10(ii)

 

Statistics – Exercise – 32.4 – Q.11

We have,

Statistics – Exercise – 32.4 – Q.11

Dividing both the sides by n we get,

Taking square root on both the sides


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