Answer to Question #293189 in Statistics and Probability for ayanda fanie

Question

In a normal distribution, 69% of the distribution is less than 28 and 90% is less than 35. Find the mean

and standard deviation of the distribution

Answer 1

We are given that,

$p(Z\lt {28-\mu\over \sigma})=0.69$ and $p(Z\lt {35-\mu\over \sigma})=0.90$

So,

$p(Z\lt {28-\mu\over \sigma})=0.69\implies \phi(0.69)={28-\mu\over \sigma}$. From the standard normal tables, we have that,

$\phi(0.69)=0.4958503\implies {28-\mu\over \sigma}=0.4958503....(1)$  

and

$p(Z\lt {35-\mu\over \sigma})=0.90\implies \phi(0.90)={35-\mu\over \sigma}$. From the standard normal tables, we have that $\phi(0.90)=1.281552$ . This implies that,   

${35-\mu\over \sigma}=1.281552.......(2)$

We shall find the values of $\mu$ and $\sigma$ by solving equations 1 and 2.

From equation 1,

${28-\mu\over \sigma}=0.4958503\implies 28-\mu=0.4958503\sigma\implies \sigma={28-\mu\over 0.4958503}.....(3)$

Substituting for the value of $\sigma$ from equation 3 in equation 2 we get,

$0.4958503(35-\mu)=1.281552(28-\mu)$

Simplifying this we have,

$17.3547605-0.4958503\mu=35.883456-1.281552\mu\implies0.7857017\mu=18.5286955\implies \mu=23.5823538$ From equation 3, we can find the value of $\sigma$ as,

$\sigma={28-23.5823538\over 0.4958503}=8.90923369$

Therefore, the values for $\mu$ and $\sigma$ are 23.58 and 8.91 respectively. Both rounded off to 2 decimal places.