Almost all medical schools in the United States require students to take the Medical College Admission Test (MCAT). To estimate the mean score 𝜇 of those who took the MCAT on your campus, you will obtain the scores of an SRS of students. The scores follow a Normal distribution, and from published information, you know that the standard deviation of scores for all MCAT takers is 10.4. Suppose that (unknown to you) the mean score of those taking the MCAT on your campus is 510.
a) Using the same sample size of 25 students, what is the probability that the mean score of your sample is between 505 and 515?
Give your answer to four decimal places.
b)If you choose one student at random, what is the probability that the student's score is between 505 and 515?
a) Z = (score-mean)/SD/√n
Look in the back of your statistics textbook for a table called something like “area under normal distribution” to find the proportions/probabilities related to the Z scores.
b) Z = (score-mean)/SD
To solve this problem, we can use the properties of the normal distribution.
a) To find the probability that the mean score of your sample is between 505 and 515, we'll use the concept of the standard error of the mean.
The standard error of the mean (SE) is calculated by dividing the standard deviation (σ) by the square root of the sample size (n). In this case, the standard error of the mean can be calculated as: SE = 10.4 / √25 = 2.08.
Next, we'll calculate the z-scores for the lower and upper bounds of the desired range using the formula: z = (x - 𝜇) / SE.
For the lower bound, z = (505 - 510) / 2.08 = -2.40.
For the upper bound, z = (515 - 510) / 2.08 = 2.40.
Using a standard normal distribution table or a calculator, we can find the probability corresponding to the z-scores -2.40 and 2.40. The probability calculated between these z-scores represents the probability that the mean score of your sample is between 505 and 515.
Therefore, the probability is P(-2.40 < z < 2.40).
b) If you choose one student at random, the probability that the student's score is between 505 and 515 can be calculated using the properties of the normal distribution and the z-score formula.
First, calculate the z-scores for the lower and upper bounds using the formula: z = (x - 𝜇) / σ, where x is the desired score, 𝜇 is the mean, and σ is the standard deviation.
For the lower bound, z = (505 - 510) / 10.4 = -0.48.
For the upper bound, z = (515 - 510) / 10.4 = 0.48.
Using a standard normal distribution table or a calculator, we can find the probability corresponding to the z-scores -0.48 and 0.48. The probability calculated between these z-scores represents the probability that a randomly selected student's score is between 505 and 515.
Therefore, the probability is P(-0.48 < z < 0.48).
Please note that in both cases, it is assumed that scores follow a normal distribution and that the sample is a simple random sample (SRS) from the population of interest.