Answer to Question #216688 in Statistics and Probability for NBa

Question

In a normal distribution 35% of the items are under 40 and 10% are over 70. Find the mean and standard deviation.

Answer 1

$P(Z<z_1)=0.35\to z_1=-0.3853.$

$P(Z>z_2)=0.10\to P(Z<z_2)=0.90\to z_2=1.2816.$

$\frac{40-\mu}{\sigma}=-0.3853.$

$\frac{70-\mu}{\sigma}=1.2816.$

$\mu-0.3853\sigma=40.$

$\mu+1.2816\sigma=70.$

$\sigma=\frac{70-40}{1.2816+0.3853}\approx18.$

$\mu=40+0.3853*18\approx47.$