Question

Almost all medical schools in the United States require students to take the Medical College Admission Test (MCAT). To estimate the mean score ???? 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 is 10.4 . Suppose that, unknown to you, the mean score of those taking the MCAT on your campus is 510 . In answering the questions, use z ‑scores rounded to two decimal places.
If you choose one student at random, what is the probability that the student's score is between 505 and 515

Answer 1

Answer:

Probability of having student's score between 505 and 515 is 0.36

Given that z-scores are rounded to two decimals using Standard Normal Distribution Table

Step-by-step explanation:

As we know from normal distribution: z(x) = (x - Mu)/SD

where x = targeted value; Mu = Mean of Normal Distribution; SD = Standard Deviation of Normal Distribution

Therefore using given data: Mu (Mean) = 510, SD = 10.4 we have z(x) by using z(x) = (x - Mu)/SD as under:

In our case, we have x = 505 & 515

Approach 1 using Standard Normal Distribution Table:

z for x=505: z(505) = (505-510)/10.4 gives us z(505) = -0.48

z for x=515: z(515) = (515-510)/10.4 gives us z(515) = 0.48

Afterwards using Normal Distribution Tables and rounding the values to two decimals we find the probabilities as under:

P(505) using z(505) = 0.32

Similarly we have:

P(515) using z(515) = 0.68

Now we may find the probability of student's score between 505 and 515 using:

P(505 < x < 515) = P(515)-P(505) = 0.68 - 0.32 = 0.36

PS: The standard normal distribution table is being attached for reference.

Approach 2 using Excel or Google Sheets:

P(x) = norm.dist(x,Mean,SD,Commutative)

P(505) = norm.dist(505,510,10.4,1)

P(515) = norm.dist(515,510,10.4,1)

Probability of student's score between 505 and 515= P(515) - P(505) = 0.36

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