Probability of a sample mean higher than population mean

I have solved this question but I am not sure if what I did is correct:

It is found that the population mean and standard deviation for the paper are 45.292 and 18.761 respectively. Find the probability of a random sample of size 250 giving a sample mean at least as high as the one found in the sample above (47.488)

The sample mean above = 47.488
simple size = 250
Sample size as high as the one found, therefor one tailed Z table (upper)

I calculated Z score as:

z_score = (xbar−mu)/(sigma/sqrt(n))
z_score = (47.488 - 45.292)/(18.761/sqrt(250))
z_score = 1.850744



With this value, I calculated the probability with upper tailed Z table to be 96.79%

I just want to check if my rationale behind the example is correct.

• A question about probability is just that: a probability calculation. The reference to "confidence level," although it is related, is irrelevant. Ultimately, the correctness of your answer rests (at least in part) on the meaning of "it is found that:" the values 45.292 and 18.761 must therefore be descriptive statistics. What kind of dataset was used, how large was it, and what formulas were used to produce these two numbers? Most likely this is a textbook or exam question and the person who created it gave no thought to this issue, but in real problems it's a crucial one.
– whuber
Commented Jun 30, 2020 at 13:22
• Hi Whuber, this is just an example in a book. We are assuming the 45.292 and 18.761 are the population parameters, and from there we are calculating the probability. Commented Jun 30, 2020 at 13:28

This is the basic idea underlying the z-test (which is correct for a sample size equal to 250 and known variance).

For the central limit theorem, you have $$\bar{x}|\mu,\sigma^2 \sim N(\mu, \sigma^2/N)$$ for $$N \rightarrow +\infty$$

which means that (considering $$N=250$$ large enough)

$$z(\bar{x})=(\bar{x}-\mu)/(\sigma/\sqrt{N})\sim N(0,1)$$

you can calculate the probability $$P(\bar{x}>47.488)=1-\Phi_Z(z(47.488))$$ where $$\Phi_Z$$ is the standard normal cumulative distribution function.

• Thank you so much Ping. I have also realised that I need to subtract [1 - 0.9679] to calculate P > 47.488, as you stated above. I really appreciate your answer. I just wanted to be sure. Best regards. Commented Jun 30, 2020 at 17:30
• As an exercise you may want to compare with the cumulative of the T-distribution. Best wishes
– user289381
Commented Jun 30, 2020 at 17:34