Question

Suppose that the random variable X has the discrete uniform distribution f(x) = { 1/4, r= 5, 6, 7, 8. 0, otherwise. A random sample of n = 45 is selected from this distribution. Find the probability that the sample mean is greater than 6.7. Round your answer to two decimal places (e.g. 98.76). P= the absolute tolerance is +/-0.01

Answer 1

Solution:

x	P(X)	x2	x*P(X)	x2*P(X)
5	1/4	25	5/4	25/4
6	1/4	36	6/4	36/4
7	1/4	49	7/4	49/4
8	1/4	64	8/4	64/4
Sum	1	-	26/4 = 6.5	174/4 = 43.5

$\mu$ = summation of x*P(X) = 6.5

Now , E(X²) = summation [x² * P(X)] = 43.5

Variance = E(X²) - [E(x)]²

= 43.5 - [6.5]²

= 1.25

Now , SD $\sigma$ = $\sqrt{}$1.25 = 1.11803399

Now sample of size 45 is selected.

n = 45

Let $url=https://latex.codecogs.com/gif.latex?%5Cbar%20x&t=1573406737&ymreqid=50870fb5-0ab3-84d7-1ca9-bb001d01e500&sig=mV2YjzkRFrqEh3lBQGaCHg--~C$ be the mean of sample.

The sampling distribution of the $url=https://latex.codecogs.com/gif.latex?%5Cbar%20x&t=1573406737&ymreqid=50870fb5-0ab3-84d7-1ca9-bb001d01e500&sig=mV2YjzkRFrqEh3lBQGaCHg--~C$ is approximately normal with

Mean($\mu_{\bar x}$) = $\mu$ = 6.5

SD($\sigma_{\bar x}$) =    = 1.11803399/$\sqrt{}$​45 = 0.1667

Find P($\bar x$ > 6.7)

= P[($\bar x$ - $\mu_{\bar x}$ )/$\sigma_{\bar x}$ >  (6.7 - $\mu_{\bar x}$ )/$\sigma_{\bar x}$]

= P[Z > (6.7 - 6.5)/0.1667 ]

= P[Z > 1.20]

= 1 - P[Z < 1.20]

= 1 - 0.8849 ( use z table)

= 0.1151 = 0.12

Answer: 0.1151 nearly 0.12