Question

In a distribution exactly normal 7%of the items are under 35 and 89% are under 63 what is the mean and standard deviation

Answer 1

Answer:

(63+35)/(89%+7%)

=98/96

=49/48

Answer 2

The mean and standard deviation are as follow:----

Mean= 50.2916 and

Std. Deviation=10.3615

Let the mean and standard deviation be ‘m′ and ′d′

Let SNV corresponding to x=35 be z₁

P(z<Z₁)=0.07

0.5– Area between ‘z=0′  to  ‘z=−z₁‘  is  0.07

Area between ‘z=0′  to  ‘z=−z₁‘  is  0.43

From  z–table, z₁=1.4758

Z₁=−1.4758

But,

z=$\frac{x-m}{d}$

z₁=$\frac{35-m}{d}$

−1.4758=$\frac{35-m}{d}$

$\frac{m-1.4758}{d}$=35          ....(1)

Let SNV corresponding to x=63 be z₂

P(x<63)=89%

P(z<Z₂)=0.89

0.5+ Area between ‘z=0′  to  ‘z=z₂′  is  0.39

Area between ‘z=0′  to  ‘z=z₂′  is  0.39

From z-table, z₂=1.2265

But Z₂=$\frac{63-m}{d}$

1.2265=$\frac{63-m}{d}$

m+1.2265d=63          …..(2)

Solving (1) and (2), we get

m=50.2916  And d=10.3615