In a normal distribution 31% of the articles are under 45 and 8% are over 64. Calculate the mean and standard deviation of the distribution.

Mean = 50, Standard Deviation = 10

The correct answer is Option (2) → Mean = 50, Standard Deviation = 10

Let X be a normal distribution random variable, then

$P(X <45) = 31\%$ i.e. 0.31

and $P(X>64) = 8\%$ i.e. 0.08

Let the mean and the standard deviation of the distribution be μ and o respectively, then

for $X = 45, Z =\frac{45-μ}{σ}$ and

for $X = 64, Z =\frac{64-μ}{σ}$

$∴P(X <45) = P\left(Z<\frac{45-μ}{σ}\right)=0.31$

$⇒P\left(Z<\frac{45-μ}{σ}\right)= P(Z < -0.5)$   (using table)

$⇒\frac{45-μ}{σ} = -0.5$   ...(i)

and $P(X >64) = P(Z > \frac{64-μ}{σ}) = 0.08=1-0.92$

$⇒P (Z > \frac{64-μ}{σ}) = 1 − P(Z < 1.4) = P(Z> 1.4)$   (using table)

$ ⇒\frac{64-μ}{σ} = 1.4$   ...(ii)

Dividing equation (i) by (ii), we get

$\frac{45 - μ}{64-μ}=\frac{-0.5}{1.4}⇒63-1.4μ=-32 +0.5μ$

$⇒ 95 = 1.9μ⇒ μ = 50$.

Substituting $μ = 50$ in equation (i), we get

$\frac{45-50}{σ}=-0.5⇒σ = 10$.

Hence, mean 50 and standard deviation = $σ = 10$.