Answer to Question #266807 in Statistics and Probability for help

Question

1.   A random sample of 16 observations is to be drawn from a normal distribution having mean 11 and standard deviation 3.  Let  denote the sample mean.  Find, correct to three decimal places

(i)         The probability that  will have a value between 9.2 and 12.2

(ii)       The value of c for which 

Answer 1

Solution:

(a) Given that,

mean $=\mu=11$

standard deviation $=\sigma=3$

$n=16 \\\mu_{\bar{x}}=11 \\\sigma_{\bar{x}}=\sigma / \sqrt n=3 / 4=0.75 \\ \left.\mathrm{P}(9.2<\bar{x}<12.2)=\mathrm{P}\left[(9.2-11) / 0.75<\left(\bar{x}-\mu_{\bar{x}}\right) / \sigma_{\bar{x}}<(12.2-11) / 0.75\right)\right] \\=\mathrm{P}(-2.4<Z<1.6) \\=\mathrm{P}(Z<1.6)-\mathrm{P}(Z<-2.4)$

Using z table

=0.9452-0.0082

probabilty =0.9370

(b) To find The value of c for which $P(\bar{x}<c)=0.03$.

$P(\bar{x}<c)=0.03$

Using standard normal table,

$\begin{aligned} &P(Z<z)=0.03 \\ &=P(Z<-1.88)=0.03 \end{aligned}$

z = -1.88 Using standard normal z table,

Using z-score formula

$\begin{aligned} &\bar{x}=\mathrm{z} \times \sigma_{\bar{x}}{+} \mu_{\bar{x}} \\ &\bar{x}=-1.88 \times 0.75+11 \\ &\bar{x}=9.59 \end{aligned}$

So, the value of c =9.59