Question

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

Answer 1

Answer on Question #45194 – Math - Statistics and Probability

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

Solution

$$Z = \frac {X - \bar {X}}{\sigma}$$

Value of $Z$, corresponding to $0.50 - 0.31 = 0.19$ area, is equal to $-0.5$ (from table).

$$- 0.5 = \frac {45 - \bar {X}}{\sigma} \rightarrow - 0.5\sigma = 45 - \bar {X} \rightarrow \bar {X} - 0.5\sigma = 45$$

Value of $Z$, corresponding to $0.5 - 0.08 = 0.42$ area, is equal to $+1.41$ (from table).

$$1.41 = \frac {64 - \bar {X}}{\sigma} \rightarrow 1.41\sigma = 64 - \bar {X} \rightarrow \bar {X} + 1.41\sigma = 64$$

Solving the system of equations

$$\left\{\begin{array}{l}\bar {X} - 0.5\sigma = 45\\\bar {X} + 1.41\sigma = 64\end{array}\right. \rightarrow - 1.91\sigma = - 19 \rightarrow \sigma = 10 \text{ approx.}$$

Substituting the value of $\sigma$ in the first equation

$$\bar {X} - 0.5 \cdot 10 = 45 \rightarrow \bar {X} = 50$$

Answer: $\bar{X} = 50$, $\sigma = 10$.

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