Question

A simple random sample of size nequals10 is obtained from a population with muequals68 and sigmaequals15. ​(a) What must be true regarding the distribution of the population in order to use the normal model to compute probabilities involving the sample​ mean? Assuming that this condition is​ true, describe the sampling distribution of x overbar. ​(b) Assuming the normal model can be​ used, determine ​P(x overbarless than71.4​). ​(c) Assuming the normal model can be​ used, determine ​P(x overbargreater than or equals69.1​).

Answer 1

(a) The distribution of the sample mean (
`\bar x`) is N (68, 4.74²).

(b) The value of
`P(\bar X<71.4)` is 0.7642.

(c) The value of
`P(\bar X\geq 69.1)` is 0.3670.

Explanation:

A random sample of size n = 10 is selected from a population.

Let the population be made up of the random variable X.

The mean and standard deviation of X are:

`\mu=68\\\sigma=15`

(a)

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Since the sample selected is not large, i.e. n = 10 < 30, for the distribution of the sample mean will be approximately normally distributed, the population from which the sample is selected must be normally distributed.

Then, the mean of the distribution of the sample mean is given by,

`\mu_(\bar x)=\mu=68`

And the standard deviation of the distribution of the sample mean is given by,

`\sigma_(\bar x)=(\sigma)/(√(n))=(15)/(√(10))=4.74`

Thus, the distribution of the sample mean (
`\bar x`) is N (68, 4.74²).

(b)

Compute the value of
`P(\bar X<71.4)` as follows:

`P(\bar X<71.4)=P((\bar X-\mu_(\bar x))/(\sigma_(\bar x))<(71.4-68)/(4.74))`

`=P(Z<0.72)\\=0.76424\\\approx0.7642`

*Use a z-table for the probability.

Thus, the value of
`P(\bar X<71.4)` is 0.7642.

(c)

Compute the value of
`P(\bar X\geq 69.1)` as follows:

Apply continuity correction as follows:

`P(\bar X\geq 69.1)=P(\bar X> 69.1+0.5)`

`=P(\bar X>69.6)`

`=P((\bar X-\mu_(\bar x))/(\sigma_(\bar x))>(69.6-68)/(4.74))`

`=P(Z>0.34)\\=1-P(Z<0.34)\\=1-0.63307\\=0.36693\\\approx0.3670`

Thus, the value of
`P(\bar X\geq 69.1)` is 0.3670.