Question

In a large lecture class, the professor announced that the scores on a recent exam were normally distributed with a range from 51 to 87. Using the Empirical Rule μ ± 3σ to estimate σ, how many students would you need to sample to estimate the true mean score for the class with 90 percent confidence and an error of ±2?

Answer 1

Answer:

The standard deviation of the scores on a recent exam is 6.

The sample size required is 25.

Step-by-step explanation:

Let X = scores of students on a recent exam.

It is provided that the random variable X is normally distributed.

According to the Empirical rule, 99.7% of the normal distribution is contained in the range, μ ± 3σ.

That is, P (μ - 3σ < X < μ + 3σ) = 0.997.

It is provided that the scores on a recent exam were normally distributed with a range from 51 to 87.

This implies that:

P (51 < X < 87) = 0.997

So,

μ - 3σ = 51...(i)

μ + 3σ = 87...(ii)

Subtract (i) and (ii) to compute the value of σ as follows:

  μ -     3σ =    51

(-)μ + (-)3σ = (-)87

______________

-6σ = -36

σ = 6

Thus, the standard deviation of the scores on a recent exam is 6.

The (1 - α)% confidence interval for population mean is given by:

$CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

The margin of error of this interval is:

$MOE = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

Given:

MOE = 2

σ = 6

Confidence level = 90%

Compute the z-score for 90% confidence level as follows:

$z_{\alpha/2}=z_{0.10/2}=z_{0.05}=1.645$

*Use a z-table.

Compute the sample required as follows:

$MOE = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\\2=1.645\times \frac{6}{\sqrt{n}}\\n=(\frac{1.645\times 6}{2})^{2}\\n=24.354225\\n\approx 25$

Thus, the sample size required is 25.