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Let {x}_{{1}},{x}_{{2}},,{x}_{{n}} be {x}_{{n}} observations.

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Let $x_{1}, x_{2},, x_{n}$ be $x_{n}$ observations. Let $y_{i} = a x_{i} + b$ for i=1,2,...,n where a and b are constants. If the mean of $x_{i}^{'} s$ is 48 and their standard deviation is 12, the mean of $y_{i}^{'} s$ is 55 and standard deviation of $y_{i}^{'} s$ is 15, the values of a and b are (a) $a = 1.25, b = - 5$ (b) $a = - 1.25, b = 5$ (c) $a = 2.5, b = - 5$ (d) $a = 2.5, b = 5$

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Harper Discussed

Let

x_{1}, x_{2},, x_{n}

x_{n}

observations. Let

y_{i} = a x_{i} + b

for i=1,2,...,n where a and b are constants. If the mean of

x_{i}^{'} s

is 48 and their standard deviation is 12, the mean of

y_{i}^{'} s

is 55 and standard deviation of

y_{i}^{'} s

is 15, the values of a and b are (a)

a = 1.25, b = - 5

(b)

a = - 1.25, b = 5

(c)

a = 2.5, b = - 5

(d)

a = 2.5, b = 5

6 mins ago

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Text solutionVerified

Solution: It is given that ,

y_{i} = a x_{i} + b

\overline{X} = 48, σ_{x} = 12

\overline{Y} = 55 and σ_{y} = 15

Now, as ,

y_{i} = a x_{i} + b

∴ \overline{Y} = a \overline{X} + b

\Rightarrow 55 = 48 a + b \to (1)

Now, their variances will also be equal.

∴ V a r (y) = V a r (a x + b)

\Rightarrow V a r (y) = a^{2} V a r (x)

...[As Variance of constants is

0

]

\Rightarrow (σ_{y})^{2} = a^{2} (σ_{x})^{2}

\Rightarrow 15^{2} = a^{2} (12)^{2}

\Rightarrow a^{2} = \frac{225}{144}

\Rightarrow a = \frac{15}{12} = \frac{5}{4}

Putting value of

a

in (1),

55 = 48 (\frac{5}{4}) + b

\Rightarrow b = 55 - 60 = - 5

∴ a = 1.25 and b = - 5

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Question Text	Let $x_{1}, x_{2},, x_{n}$ be $x_{n}$ observations. Let $y_{i} = a x_{i} + b$ for i=1,2,...,n where a and b are constants. If the mean of $x_{i}^{'} s$ is 48 and their standard deviation is 12, the mean of $y_{i}^{'} s$ is 55 and standard deviation of $y_{i}^{'} s$ is 15, the values of a and b are (a) $a = 1.25, b = - 5$ (b) $a = - 1.25, b = 5$ (c) $a = 2.5, b = - 5$ (d) $a = 2.5, b = 5$
Answer Type	Text solution:1
Upvotes	150