Chapter 32: Statistics – Exercise 32.4

Statistics – Exercise – 32.4 – Q.1(i)

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

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

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.

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 Σx_i = Incorrect Σx_i - (Sum of incorrect values) + (Sum of correct values)

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

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

891

Statistics – Exercise – 32.4 – Q.8

We have,

n = 100,= 40, σ = 5.1

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

= 4000 - 50 + 40 = 3990

= 162601 - (50)² + (40)² = 161701

Statistics – Exercise – 32.4 – Q.9

We have, 

n = 20,= 10 and σ = 2

⟹ Incorrected Σx_i = 200

and,

(i) When 8 is omitted from the data:

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

Now, Incorrected Σx_i = 200

⟹ Corrected Σx_i + 8 = 200

⟹ Corrected Σx_i = 192

and, 

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

we have, Incorrected Σx_i = 200

∴ Corrected Σx_i = 200 - 8 + 12 = 204

and,

Statistics – Exercise – 32.4 – Q.10

We have, 

⟹ Incorrect Σx_i = 2000

and,

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

n = 97

Now, Incorrect Σx_i = 2000

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

and, 

Statistics – Exercise – 32.4 – Q.11

We have,

Dividing both the sides by n we get,

Taking square root on both the sides