Question

To estimate the mean score μ of those who took the Medical College Admission Test on your campus, you will obtain the scores of an SRS of students. From published information you know that the scores are approximately Normal with standard deviation about 6.5. You want your sample mean x¯ to estimate μ with an error of no more than 1.3 point in either direction.

(a) What standard deviation (±0.0001) must x¯ have so that 99.7% of all samples give an x¯ within 1.3 point of μ?

(b) How large an SRS do you need in order to reduce the standard deviation of x¯ to the value you found in part (a)?

Answer 1

Answer:

Step-by-step explanation:

Let X be the SRS scores of students.

Given that X is normal with standard deviation about 6.5.

Mean difference = $|bar x-\mu|<1.3$ is desired

i.e. Z score should be $|z|<\frac{1.3}{s} =0.997$

Std error = std deviation of sample = 0.447

b) $z =2.95\\\frac{1.3}{s} =2.95\\s = 0.447\\6.5/\sqrt{n} =0.447\\n >193$

So std error must be 0.997 and sample size must be atleast 193 to get the difference of x bar and mu to lie within 1.3 points in either direction

Answer 2

Using the normal distribution and the central limit theorem, it is found that:

a) A standard deviation of 0.4333 would be required.

b) An SRS of 226 would be needed.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.  
• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
• By the Central Limit Theorem, for samples of size n, the standard deviation is of $s = \frac{\sigma}{\sqrt{n}}$

Item a:

• By the Empirical Rule, 99.7% of the measures are within 3 standard deviations of the mean.

• Thus, for it to be within 1.3 points of the mean, we want to find $\sigma$ for which: $X - \mu = 1.3, Z = 3$

Then:

$Z = \frac{X - \mu}{\sigma}$

$3 = \frac{1.3}{\sigma}$

$3\sigma = 1.3$

$\sigma = \frac{1.3}{3}$

$\sigma = 0.4333$

A standard deviation of 0.4333 would be required.

Item b:

For the distribution, we have that $\sigma = 6.5$.

We want to find n for which $s = 0.4333$. Thus:

$s = \frac{\sigma}{\sqrt{n}}$

$0.4333 = \frac{6.5}{\sqrt{n}}$

$0.4333\sqrt{n} = 6.5$

$\sqrt{n} = \frac{6.5}{0.4333}$

$(\sqrt{n})^2 = (\frac{6.5}{0.4333})^2$

$n = 225.05$

Rounding up:

An SRS of 226 would be needed.

A similar problem is given at https://brainly.com/question/24663213