Answer to Question #191688 in Statistics and Probability for Miras Argynbay

Question

The set X = {12, 15, a, b} has a mean 15.5 and median 16. It is given that set X is arranged in ascending order.

Determine the values of a and b. [2 marks]

Find the standard deviation of set X. [1 mark]

When each number in set X is decreased by 10, find the mean and standard deviation. [2 marks]

If each number of X is increased by 100%, find the mean and standard deviation.

Answer 1

The given set is-

X={12,15,a,b}

$\text{Mean} =\dfrac{12+15+a+b}{4} \\[9pt] 15.5=\dfrac{27+a+b}{4} \\[9pt] \Rightarrow a+b=15.5\times 4-27 \\[9pt] \Rightarrow a+b=35~~~~~-(1)$

$Median =(\dfrac{N}{2}+1)^{th} term \\[9pt] 16=(\dfrac{4}{2}+1)^{th} term \\[9pt] 16=3^{th} term$

hence a$=16$ and $b=35-16=19$

Now The set X is {12,15,16,19}

Variance $=\dfrac{\sum(x_i-\bar{x})^2}{4}=\dfrac{(-3.5)^2+(-0.5)^2+(0.5)^2+(3.5)^2}{4}=\dfrac{2405.5}{4}=612.625$

Standard deviation $=\sqrt{\text{variance}}=\sqrt{612.625}=24.75$

When each number is decreased by 10 , The new set became {2,5,6,9}

New Mean $=\dfrac{2+5+6+9}{4}=\dfrac{22}{4}=5.5$

Standard deviation remains same i.e. 24.75

When each number is increased by 100%, The new set is {24,30,32,38}

New mean $=\dfrac{24+30+32+38}{4}=\dfrac{124}{4}=31$

Standard deviation remains same as 24.75