Question

Almost all medical schools in the United States require students to take the Medical College Admission Test (MCAT). To estimate the mean score μμ 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 6.4. Suppose that (unknown to you) the mean score of those taking the MCAT on your campus is 24.

In answering the following, use z-scores rounded to two decimal places.

If you choose one student at random, what is the probability (±±0.0001) that the student's score is between 20 and 30?

You sample 22 students. What is the standard deviation (±±0.01) of sampling distribution of their average score x¯¯¯x¯?

What is the probability (±±0.0001) that the mean score of your sample is between 20 and 30?

Answer 1

Answer:

a) 0.5588

b) 0.9984

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 24

Standard Deviation, σ = 6.4

We are given that the distribution of score is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$        

a) P(score between 20 and 30)

$P(20 \leq x \leq 30) = P(\displaystyle\frac{20 - 24}{6.4} \leq z \leq \displaystyle\frac{30-24}{6.4}) = P(-0.62 \leq z \leq 0.94)\\\\= P(z \leq 0.94) - P(z < -0.62)\\= 0.8264 - 0.2676 = 0.5588 = 55.88\%$

$P(20 \leq x \leq 30) = 55.88\%$

b) Sampling distribution

Sample size, n = 22

The sample will follow a normal distribution with mean 24 and standard deviation,

$s = \dfrac{\sigma}{\sqrt{n}} = \dfrac{6.4}{\sqrt{22}} =1.36$

c) P(mean score of sample is between 20 and 30)

$P(20 \leq x \leq 30) = P(\displaystyle\frac{20 - 24}{1.36} \leq z \leq \displaystyle\frac{30-24}{1.36}) = P(-2.94 \leq z \leq 4.41)\\\\= P(z \leq 4.41) - P(z < -2.94)\\= 1 - 0.0016 = 0.9984 = 99.84\%$

$P(20 \leq x \leq 30) =99.84\%$

Answer 2

The probability that the mean score of your sample is between 20 and 30 is 99.84%.

What is a Z-table?

A z-table also known as the standard normal distribution table, helps us to know the percentage of values that are below (or to the left of the Distribution) a z-score in the standard normal distribution.

As it is given that the mean(μ) is 24, while the standard deviation is 6.4. Also, it is given that the distribution of score is a bell-shaped distribution therefore, it is a normal distribution.

Since the distribution is normal distribution, therefore, we will use a z-score,

A.) P(score between 20 and 30)

$\begin{aligned}P(20\leq x\leq 30) &= P(\dfrac{20-24}{6.4}\leq z\leq \dfrac{30-24}{6.4})\\\\&=P(-0.62\leq z\leq 0.94)\\\\&=P(z\leq 0.94) - P(z\leq -0.62)\\\\\text{Using the Z-table}\\\\&=0.8264-0.2676\\\\&=0.5588 = 55.88\%\end{aligned}$

B.) Sampling distribution

As the sample size, n = 22, and, the sample will follow a normal distribution with a mean of 24. Therefore, the standard deviation can be written as,

$s = \dfrac{\sigma}{\sqrt n} = \dfrac{6.4}{\sqrt{22}} = 1.36$

C.) P(mean score of sample is between 20 and 30)

$\begin{aligned}P(20\leq x\leq 30) &= P(\dfrac{20-24}{1.36}\leq z\leq \dfrac{30-24}{1.36})\\\\&=P(-2.94\leq z\leq 4.41)\\\\&=P(z\leq 4.41) - P(z\leq -2.94)\\\\\text{Using the Z-table}\\\\&=1-0.0016\\\\&=0.9984= 99.84\%\end{aligned}$

Hence, the probability that the mean score of your sample is between 20 and 30 is 99.84%.

Learn more about Z-table:

https://brainly.com/question/6096474