The mean of five numbers 3, 5, 7, x, and y is 6. Given that the variance of the numbers is… — JAMB UTME

The mean of five numbers $3, 5, 7, x,$ and $y$ is $6$ . Given that the variance of the numbers is $8$ and $x < y$ , calculate the value of $y$ .

Cevap: 11

Cevap

The total sum of the five numbers is

5 \times 6 = 30

, giving

x + y = 15

. The sum of squared deviations from the mean

6

5 \times 8 = 40

. The known numbers

3, 5, 7

contribute

(-3)^2 + (-1)^2 + 1^2 = 11

to this sum, leaving

(x-6)^2 + (y-6)^2 = 29

. Substituting

y = 15 - x

yields

2x^2 - 30x + 88 = 0 \implies x^2 - 15x + 44 = 0

. Factoring gives roots

4

and

11

. Since

x < y

, we find

y = 11

Adım Adım Çözüm

Use the definition of the arithmetic mean to write a linear relationship between

x

and

y

x + y = 15

, or

y = 15 - x

The total sum of 5 numbers with a mean of 6 is

5 \times 6 = 30

. Subtracting the known numbers

3 + 5 + 7 = 15

leaves

x + y = 15

Apply the variance formula for population data.

\sum (x_i - 6)^2 = 40

Variance is the mean of squared deviations from the mean:

\frac{\sum (x_i - \bar{x})^2}{5} = 8 \implies \sum (x_i - 6)^2 = 40

Compute the sum of squared deviations for the known elements and simplify the variance equation.

(x - 6)^2 + (y - 6)^2 = 29

The squared deviations for

3, 5, 7

are

(-3)^2 = 9

(-1)^2 = 1

, and

1^2 = 1

. Subtracting

9 + 1 + 1 = 11

from

40

gives

29

Substitute

y = 15 - x

into the simplified equation and solve the resulting quadratic equation.

x = 4

x = 11

Substituting

y = 15 - x

gives

(x - 6)^2 + (9 - x)^2 = 29 \implies 2x^2 - 30x + 88 = 0 \implies x^2 - 15x + 44 = 0 \implies (x - 4)(x - 11) = 0

Select the correct pair

(x, y)

using the condition

x < y

x = 4

and

y = 11

Since

x < y

x

must be the smaller value (

4

) and

y

must be the larger value (

11

Anahtar Kavram

Calculation of variance and mean for ungrouped data containing unknown elements

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