Question

94.5. You intend to draw a random population of values has a normal distribution with u = 88.4 and o = sample of size n = 79. Find the probability that a single randomly selected value is greater than 71.4. P(X > 71.4) = 0.4286 X Find the probability that a sample of size n = 79 is randomly selected with a mean greater than 71.4. Pī > 71.4) 0.0548 x Enter your answers as numbers accurate to 4 decimal places. Tip: Use the Desmos calculator... 7

Answer 1

Let the random variable X follow normal distribution with mean 88.4 and standard deviation 94.5

$\therefore X\sim N(88.4,94.5)$

Here sample size n=79

We need to find the probability that

= P X - 88.4 94.5 71.4 - 88.4 94.5

= P(Z > -0.17989418)

$=1-P\left ( Z\leq -0.17989418 \right )$

$=1-[1-P\left ( Z\leq 0.17989418 \right )]$

$=P\left ( Z\leq 0.17989418 \right )$

$=0.571382178$

$\simeq 0.5714$[ round to four decimal place]

The probability is 0.5714

The distribution of sample mean is given by,

$\therefore \bar X\sim N\left ( \mu,\frac{\sigma}{\sqrt n} \right )$

So, here $\therefore \bar X\sim N\left ( 88.4,\frac{94.5}{\sqrt {79}} \right )$

We need to find $P(\bar X>71.4)$

$P(\bar X>71.4)$

$=P\left ( \frac{\bar X-88.4}{\frac{94.5}{\sqrt {79}}}>\frac{71.4-88.4}{\frac{94.5}{\sqrt {79}}} \right )$

$=P\left ( Z>\frac{71.4-88.4}{\frac{94.5}{\sqrt {79}}} \right )$

$=P\left ( Z>-1.598934445 \right )$

$=1-P\left ( Z\leq -1.598934445 \right )$

$=1-[1-P\left ( Z\leq 1.598934445 \right )]$

$=P\left ( Z\leq 1.598934445 \right )$

$=0.945082415$

$\simeq 0.9451$

The probability is 0.9451