Answer to Question #122956 in Statistics and Probability for nishna jagannadhan

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

$$P(Z<z_1)=0.31=>z_1\approx-0.49585$$

$$P(Z>z_2)=0.08=>z_2\approx1.40507$$

$${45-\mu \over \sigma}=-0.49585$$

$${64-\mu \over \sigma}=1.40507$$

$${45-\mu \over -0.49585}={64-\mu \over 1.40507}$$

$$\mu={ 45(1.40507)+64(0.49585)\over 1.40507+0.49585}\approx49.9561$$

$$\sigma={64-49.9561 \over 1.40507}=9.9952$$