Subject: Applied Mathematics 4

Topic: Probability

Difficulty: Medium

Let the mean of the distribution be $\mu$ and its standard deviation be $\sigma$.

The standard normal variate $ Z = \frac{X - \mu}{\sigma} $

P(X $\leq$ 35) = 7% = 0.07

P(X $\leq$ 63) = 89% = 0.89

When X = 35, $ Z = \frac{35 - \mu}{\sigma} $

When X = 63, $ Z = \frac{63 - \mu}{\sigma} $

P(Z $\leq$ Z$_1$) = 0.07 and P(Z $\leq$ Z$_2$) = 0.89

P(Z $\leq$ Z$_1$) = 0.07 $\implies$ P(Z$_1$ $\leq$ Z $\leq$ 0) = 0.5 - 0.07 = 0.43

By Symmetry P(0 $\leq$ Z $\leq$ Z$_1$) = 0.43

$\therefore Z_1$ = -1.48 [Z = Z$_1$ and lies on the left]

P(Z $\leq$ Z$_2$) = 0.89

P(Z $\leq$ 0) + P(0 $\leq$ Z $\leq$ Z$_2$) = 0.89

0.5 + P(0 $\leq$ Z $\leq$ Z$_2$) = 0.89

P(0 $\leq$ Z $\leq$ Z$_2$) = 0.39

Z$_2$ = 1.23

Now, Z$_1$ = -1.48

$ \frac{35 - \mu}{\sigma} = -1.48 \\ \mu - 1.48 \sigma = 35 \hspace{0.5cm} ...(1) $

Also, Z$_2$ = 1.23

$ \frac{63 - \mu}{\sigma} = 1.23 \\ \mu + 1.23 \sigma = 63 \hspace{0.5cm} ...(2) $

Solving (1) and (2)

$\mu$ = 50.2915 and $\sigma$ = 10.332