Question

Suppose a random variable is normally distributed. The probabilities for a85 and 142 are 10% and 65%, respectively. Find the mean and standard deviation to the nearest whole number.

Answer 1

Let a ~ $N(b, \sigma^2)=b+\sigma N(0,1)$ , then

$P(a<85)=0.1\implies P(b+\sigma N(0,1)<85)=0.1\implies P(N(0,1)<{\frac {85-b} {\sigma}})=0.1\implies {\frac {85-b} {\sigma}}=-1.28\implies b=1.28\sigma+85(*)$

$P(a<142)=0.65\implies P(b+\sigma N(0,1)<142)=0.65\implies P(N(0,1)<{\frac {142-b} {\sigma}})=0.65\implies {\frac {142-b} {\sigma}}=0.39\implies b=142-0.39\sigma(**)$

From (*) and (**) we have that $85+1.28\sigma=142-0.39\sigma\implies \sigma=34.1\implies b=128.6$

The mean is 129, the standard deviation is 34