**1 Answer**

written 8.0 years ago by |

Let the mean and standard deviation be $‘m’$ and $'\sigma'$

Let SNV corresponding to $x = 35$ be $z_1$

$$P (x \lt 35) = 35\% $$

$ P (z \lt Z_1) = 0.07 \\ 0.5 – \text { Area between } ‘z = 0’ \space \space to\space \space ‘z = -z_1 ‘ \space \space is \space\space 0.07 \\ \text {Area between } ‘z = 0’ \space\space to \space\space ‘z = -z1 ‘\space\space is \space\space 0.43 \\ From \space\space z – table, -z_1 = 1.4758 \\ Z_1 = -1.4758 \\ But, z=\dfrac {x-m}{\sigma} \\ z_1=\dfrac {35-m}{\sigma} \\ -1.4758=\dfrac {35-8}{\sigma}\\ m-1.4758\sigma =35.... (1) $

Let SNV corresponding to $x = 63$ be $z_2 \\ P (x \lt 63) = 89\% \\ P (z \lt Z_2) = 0.89$

$0.5 +$ Area between $‘z = 0’ \space \space to\space \space ‘z = z2’ \space \space is \space \space 0.39$

Area between $‘z = 0’ \space \space to \space \space ‘z = z2’ \space \space is\space \space 0.39$

From z-table, $z_2 = 1.2265$

But $Z_2 = \dfrac {63-m}{\sigma} \\ 1.2265 = \dfrac {63-m}{\sigma} \\ M + 1.2265\sigma = 63 ….. (2) $

Solving (1) and (2), we get

$M = 50.2916 $ And $\sigma = 10.3615 $

Now Probability that an item selected at random lies between 45 & 56 = $P (45 \lt x \lt 56)\\ =P\Big[\dfrac {45-50.2916}{10.3615} \lt \dfrac {x-m}{\sigma} \lt \dfrac {56-50.2916}{10.3615}\Big]\\ = P[-0.5106 \lt z \lt 0.5509] $

= Area between $‘z = 0’ \space \space to\space \space ‘z = -0.5106’$ + Area between $‘z = 0’ \space \space to \space \space ‘z = -0.5509’ \\ = 0.9152 + 0.2092 \\ = 0.4044$

Mean $= 50.2916,$ standard deviation $= 10.3615$

Probability that an item selected at random lies between

$$45\space\space to \space\space56 = 0.4044$$