Question

Suppose x has a distribution with μ = 10 and σ = 9. (a) If a random sample of size n = 43 is drawn, find μx, σ x and P(10 ≤ x ≤ 12). (Round σx to two decimal places and the probability to four decimal places.)

Answer 1

Answer:

P(10 ≤ x ≤ 12) = 0.4274

Step-by-step explanation:

Population mean = u = 10

Population Standard Deviation = $\sigma$ = 9

Sample size = n = 43

Sample mean($\mu_{x}$) is equal to the population mean. So,

Sample mean = $\mu_{x}$ = 10

Sample standard deviation($\sigma_{x}$) is equal to population standard deviation divided by square root of sample size. So,

Sample standard deviation = $\sigma_{x}$ = $\frac{9}{\sqrt{43}}=1.372$

We have to find the probability that for a random sample of n = 43, the value lies between 10 and 12 i.e. P(10 ≤ x ≤ 12)

P(10 ≤ x ≤ 12) = P(x ≤ 12) - P( x ≤ 10)

We can find P(x ≤ 12 ) and P(x ≤ 10) by converting these values to z scores.

The formula for z score is:

$z=\frac{x-\mu_{x}}{\sigma_{x}}$

For x =12, we get:

$z=\frac{12-10}{1.3725}=1.457$

For x =10, we get:

$z=\frac{10-10}{1.3725}=0$

So,

P(x ≤ 12) - P( x ≤ 10) = P(z ≤ 1.457) - P(z ≤ 0)

From the z table,

P(z ≤ 1.457) = 0.9274

P(z ≤ 0) = 0.5

So,

P(x ≤ 12) - P( x ≤ 10) = P(z ≤ 1.458) - P(z ≤ 0) = 0.9274 - 0.5 = 0.4274

So,

P(10 ≤ x ≤ 12) = P(x ≤ 12) - P( x ≤ 10) = 0.4274

Therefore,

The probability that for a random sample of size 43, the mean lies between 10 and 12 is 0.4274.