Suppose the amount of time it takes for statistics students to complete their midterm is normally distributed with a mean of 85.70 minutes and a standard deviation of 22.55 minutes. If 10 students are randomly selected, what is the probability that they would average less than one and three-quarter hours to take their midterm?
Z = (score-mean)/SEm
SEm = SD/√(n-1)
Change all into minutes and calculate Z.
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 10 students would average less than one and three-quarter hours to take their midterm, we need to calculate the probability distribution of the sample mean.
Step 1: Convert one and three-quarter hours into minutes.
One hour is equal to 60 minutes, and three-quarters of an hour is 45 minutes. Therefore, one and three-quarter hours is equal to 60 + 45 = 105 minutes.
Step 2: Find the mean and standard deviation of the sample mean distribution.
The mean of the sample mean distribution is equal to the population mean, which is 85.70 minutes.
The standard deviation of the sample mean distribution, also known as the standard error, can be calculated by dividing the population standard deviation by the square root of the sample size. In this case, the sample size is 10.
Standard deviation of the sample mean = Standard deviation of population / √(sample size)
Standard deviation of the sample mean = 22.55 / √10 ≈ 7.13
Step 3: Calculate the z-score.
The z-score represents the number of standard deviations a particular value is from the mean. In this case, the z-score is calculated using the sample mean, the population mean, and the standard deviation of the sample mean.
z = (sample mean - population mean) / standard deviation of the sample mean
z = (105 - 85.70) / 7.13 ≈ 2.70
Step 4: Determine the probability using the z-score.
With the calculated z-score of 2.70, we can refer to a standard normal distribution table or use statistical software to find the probability associated with this z-score. The probability can be found in two ways:
a) Look up the z-score in a standard normal distribution table. The probability associated with a z-score of 2.70 is approximately 0.9965.
b) Use statistical software or a calculator to find the probability directly. The probability associated with a z-score of 2.70 is approximately 0.9965.
Therefore, the probability that 10 students would average less than one and three-quarter hours to take their midterm is approximately 0.9965, or 99.65%.