Question

Samples of certain size are drawn from a normally distributed population with S. D. 16. It is observed that the probability

of the sample mean lying between 9.8 and 14.6 is 45.14%. Find the sample-size and the population mean. (Given that

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Answer 1

The probability given is,

$p(9.8\lt\bar{x}\lt14.6)=0.4514$, with $\sigma=16$

Since the sample mean is normally distributed, we standardize this probability as follows,

$p({(9.8-\mu)\over({16\over\sqrt{n}})}\lt Z\lt {(14.6-\mu)\over({16\over\sqrt{n}})})=0.4514$

Since the the two points given are symmetric about the population mean, it implies that to find probability for each point, we divide the given probability by 2. That is, ${0.4514\over 2}=0.2257$.

Let, ${(9.8-\mu)\over({16\over\sqrt{n}})}$ be $Z_0$ and ${(14.6-\mu)\over({16\over\sqrt{n}})}$ be $Z_1$.

So,

$p(Z_0\lt Z\lt 0)=0.2257\implies\phi(0)-\phi(Z_0)=0.2257$

Since $\phi(0)=0.5,\phi(Z_0)=0.5-0.2257=0.2743$

From the standard normal tables, we can see that $Z_0=-0.6$

Also,

$p(0\lt Z\lt Z_1)=0.2257\implies \phi(Z_1)-\phi(0)=0.2257\implies \phi(Z_1)=0.2257+0.5=0.7257$

From the standard normal tables, we can see that $Z_1=0.6$

Therefore,

$Z_0={(9.8-\mu)\over({16\over\sqrt{n}})}=-0.6\implies 9.8\sqrt{n}-\mu\sqrt{n}=-9.6......(i)$

$Z_1= {(14.6-\mu)\over({16\over\sqrt{n}})}=0.6\implies14.6\sqrt{n}-\mu\sqrt{n}=9.6......(ii)$

Subtracting equation $(i)$ from equation $(ii)$ we have,

$4.8\sqrt{n}=19.2\implies\sqrt{n}={19.2\over4.8}\implies n=16$

We take any of the two equations and solve for $\mu$,

Taking equation $(i),$

$9.8*4-4\mu=-9.6\implies (39.2+9.6)=4\mu\implies \mu=12.2$

Therefore, the population mean $\mu$ and the sample size $n$ are 12.2 and 16 respectively.