Suppose a random sample of 25 students is selected from a community college where the scores in the final exam (out of 125 points) are normally distributed, with mean equal to 112 and standard deviation equal to 12. Find the probability that the sample mean deviates from the population mean ì = 112 by no more than 4.
a.0.9050
b.0.9554
c.0.8767
d.0.9914
I don't see how it could be normally distributed with max of 125, mean = 112 and SD = 12. There could be no scores above 125, which is slightly more than 1 SD above the mean.
To find the probability that the sample mean deviates from the population mean by no more than 4, we can use the Central Limit Theorem. According to the Central Limit Theorem, the distribution of sample means will approach a normal distribution, regardless of the shape of the population distribution, as the sample size increases.
First, we need to calculate the standard deviation of the sample mean. The standard deviation of the sample mean, also known as the standard error, is given by the formula:
Standard error = standard deviation / sqrt(sample size)
In this case, the standard error = 12 / sqrt(25) = 12 / 5 = 2.4.
Next, we need to standardize the deviation from the population mean by calculating the z-score using the formula:
z = (sample mean - population mean) / standard error
In this case, we want the deviation to be no more than 4. So, we have:
z = (4 - 0) / 2.4 = 1.67
Finally, we can find the probability using the standard normal distribution table. The probability corresponds to the area under the curve to the left of the z-score.
Using the standard normal distribution table, we find that the probability corresponding to a z-score of 1.67 is approximately 0.9525.
So, the answer is b. 0.9554.