The set of four numbers \{x, 6, 8, 10\} has a variance of 5. Given that x < 6, find the va… — JAMB UTME

The set of four numbers $\{x, 6, 8, 10\}$ has a variance of $5$ . Given that $x < 6$ , find the value of $x$ .

Answer: 4

The value of

x

is 4.

By expressing the mean as

\bar{x} = \frac{x+24}{4}

and setting the sum of squared deviations divided by 4 equal to 5, we arrive at the quadratic equation

x^2 - 16x + 48 = 0

. Solving this equation gives two possible values,

x = 4

and

x = 12

. Applying the restriction

x < 6

uniquely identifies

x = 4

as the correct solution.

Calculate the mean of the dataset in terms of

x

\bar{x} = \frac{x + 24}{4}

The mean of a dataset is the sum of all values divided by the total number of items.

Write the variance equation using the formula

\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{N}

\frac{(x - \bar{x})^2 + (6 - \bar{x})^2 + (8 - \bar{x})^2 + (10 - \bar{x})^2}{4} = 5

The given variance is 5 and the number of observations

N = 4

Substitute

\bar{x} = \frac{x + 24}{4}

and expand the squared deviations

\frac{3}{16}x^2 - 3x + 14 = 5 \implies x^2 - 16x + 48 = 0

Expanding the squared terms and multiplying through by

\frac{16}{3}

yields a standard quadratic equation.

Solve the quadratic equation

x^2 - 16x + 48 = 0

for

x

(x - 4)(x - 12) = 0 \implies x = 4 \text{ or } x = 12

Factoring the quadratic equation gives two possible roots.

Apply the given condition

x < 6

x = 4

Since

x < 6

, we reject

x = 12

and select

x = 4

Variance of Ungrouped Data and Algebraic Problem Solving

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