A final exam in sociology has a mean of 72 and a standard deviation of 9.2. If 35 students are randomly selected, find the probability that the mean of their test scores will be greater than 76. Round to tenth of a percent
Z = (score-mean)/SEm
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.
To find the probability that the mean of the test scores will be greater than 76, we need to use the Central Limit Theorem. According to the Central Limit Theorem, the sampling distribution of the sample means will approach a normal distribution, regardless of the shape of the population distribution, as the sample size increases.
The mean of the sampling distribution of the sample means is equal to the population mean, which is 72 in this case. The standard deviation of the sampling distribution of the sample means, also known as the standard error, is equal to the population standard deviation divided by the square root of the sample size.
First, we need to calculate the standard error:
Standard Error = Standard Deviation / √Sample Size = 9.2 / √35 ≈ 1.556
Next, we calculate the z-score, which is the number of standard errors that the sample mean is away from the population mean:
z = (Sample Mean - Population Mean) / Standard Error
= (76 - 72) / 1.556
= 2.571
To find the probability corresponding to the z-score of 2.571, we can consult a standard normal distribution table or use statistical software.
Using a standard normal distribution table or statistical software, we find that the probability of getting a z-score greater than 2.571 is approximately 0.0058.
To express this probability as a percentage rounded to the tenth, we multiply by 100 and round to the nearest tenth:
0.0058 * 100 ≈ 0.6%
Therefore, the probability that the mean of the test scores will be greater than 76 is approximately 0.6%.