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 μ = 112 by no more than 4.
Z = (score-mean)/SD
Solve for scores of 116 and 108.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability between the two Z scores.
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. The Central Limit Theorem states that the distribution of sample means approaches a normal distribution as the sample size increases, regardless of the shape of the population distribution.
First, we need to calculate the standard error of the sample mean, which is the standard deviation of the population divided by the square root of the sample size.
Standard Error (SE) = Standard Deviation (σ) / √Sample Size (n)
SE = 12 / √25
SE = 12 / 5
SE = 2.4
Next, we need to find the z-score for a deviation of 4 from the population mean. The z-score represents the number of standard errors a particular value is away from the mean. The formula for z-score is (x - μ) / SE, where x is the value, μ is the population mean, and SE is the standard error.
z-score = (4 - 0) / 2.4
z-score = 4 / 2.4
z-score = 1.67
Using a standard normal distribution table or a statistical calculator, we can find the probability associated with the z-score of 1.67. The probability is the area under the normal distribution curve to the left of the z-score.
The probability P(Z < 1.67) is approximately 0.9525.
However, we want the probability that the sample mean deviates from the population mean by no more than 4, so we need to find the probability for both positive and negative deviations. Since the distribution is symmetrical, we can multiply the probability by 2.
Probability = 2 * 0.9525
Probability = 1.905
Therefore, the probability that the sample mean deviates from the population mean by no more than 4 is approximately 1.905 (or 19.05%).
To find the probability that the sample mean deviates from the population mean μ by no more than 4, we need to use the concept of the sampling distribution of the mean.
The mean of the sampling distribution of the mean is the same as the population mean μ. In this case, μ = 112.
The standard deviation of the sampling distribution of the mean, also known as the standard error, can be calculated using the formula:
Standard Error = standard deviation / √sample size
In this case, the standard deviation is 12 and the sample size is 25. Therefore, the standard error can be calculated as:
Standard Error = 12 / √25 = 12 / 5 = 2.4
Now, we need to find the probability that the sample mean deviates from the population mean by no more than 4 standard errors. Since we have a normal distribution, we can use the Z-score formula to standardize the deviation.
Z-score = (X - μ) / standard error
In this case, X represents the value we are interested in, which is 4 standard errors. So, X = 4 * 2.4 = 9.6.
Z-score = (9.6 - 0) / 2.4 = 4
Now, we need to find the probability of Z-score ≤ 4. We can use a Z-table or a calculator to look up this probability. Looking it up, we find that the probability of Z-score ≤ 4 is approximately 0.9999.
Therefore, the probability that the sample mean deviates from the population mean μ = 112 by no more than 4 is approximately 0.9999 or 99.99%.