Question

IQ test had a mean score of 100 with a standard deviation of 15. Assume that the

scores are normally distributed.

1. What is the lowest score that would still place a student in the top 5% of the

scores?

2. What is the highest score that would still place a student in the bottom 10%

of the scores?

3. A random sample of 60 students is drawn from this population. What is the

probability that the mean IQ score is greater than 105? Interpret your result.

4. Are you more likely to randomly select one student with an IQ score greater

than 105 or are you more likely to randomly select a sample of 15 students

with a mean IQ score greater than 105? Explain.

Answer 1

Q1)

Our objective is to find the lowest score that would still place a student in the top \(5 \%\) of the scores.

A test score in the top \(5 \%\) is any score above the \(95^{\text {th }}\) percentile. To find the score that represents the \(95^{\text {th }}\) percentile, we must find the \(z\) -score that corresponds to a cumulative area of

0.95.

From the Standard Normal Table, we can find that the area closest to 0.95 in the table are0.9495

\((z=1.64)\) and \(0.9505(z=1.65)\)

Because 0.95 is halfway between the two areas in the table, use the z-score that is halfway between 1.64 and 1.65 .

So, the z-score that corresponds to an area of 0.95 is \(1.645 .\)

Student IQ score in the top \(5 \%\) correspond to the shaded region shown below.

Using the equation \(x=\mu+z \sigma,\) we get the required score.

$$ x=\mu+z \sigma $$

\(=100+1.645(15)\)

$$ \begin{array}{l} =100+24.675 \\ \approx 124.68 \end{array} $$

Therefore, the lowest score that would still place a student in the top \(5 \%\) of the scores is 124.68.

Answer 2

Q3)

The sample size (60) is greater than \(30,\) so we can use the Central limit theorem to conclude that the distribution of the sample means is approximately normal, with mean and standard deviations

$$ \mu_{\bar{x}}=\mu \text { and } \sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}} \text { respectively. } $$

Mean of the sampling distribution of the means, \(\mu_{\bar{x}}=\mu\)

$$ =100 $$

Standard deviation of the sampling distribution of means, \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\)

$$ \begin{array}{l} =\frac{15}{\sqrt{60}} \\ \approx 1.936 \end{array} $$

The z-score that corresponds to 105 is shown below.

$$ \begin{aligned} z &=\frac{\bar{x}-\mu}{\sigma / \sqrt{n}} \\ &=\frac{105-100}{15 / \sqrt{60}} \\ &=\frac{5}{1.936} \\ & \approx 2.58 \end{aligned} $$

Probability that the mean IQ score is greater than 105 is shown below.

$$ P(\bar{x}>105)=P(z>2.58) $$

$$ \begin{array}{l} =1-P(z<2.58) \\ =1-0.9951 \quad\left(\begin{array}{l} \text { Using Excel function } \\ "=\text { NORMSDIST }(2.58)^{\prime \prime} \end{array}\right) \\ =0.0049 \end{array} $$

Therefore, \(0.49 \%\) of such samples with \(n=60\) will have a mean greater than 105 and largely \(99.51 \%\) of these sample means will lie outside this interval.

If the probability of an event is less than \(0.05,\) we consider those events as unusual events.

Here the event "mean IQ score is greater than 105 for a sample of size \(60^{\prime \prime}\) is unusual

because its probability is less than \(0.05 .\)

Answer 3

Q4)

Probability that a student had an IQ score greater than 105 is obtained below.

$$ \begin{aligned} P(x>105) &=P\left(\frac{x-\mu}{\sigma}>\frac{105-100}{15}\right) \\ &=P(z>0.33) \\ &=1-P(z<0.33) \\ &=1-0.6293 \quad\left(\begin{array}{l} \text { Using Excel function } \\ \text { "NORMSDIST(0.33)" } \end{array}\right) \\ =0.3707 \end{aligned} $$

Suppose a random sample of 15 students is drawn this population.

Probability that the mean IQ score greater than 105 is obtained below.

$$ \begin{aligned} P(\bar{x}>105) &=P\left(\frac{\bar{x}-\mu}{\sigma / \sqrt{n}}>\frac{105-100}{15 / \sqrt{15}}\right) \\ &=P(z>1.29) \end{aligned} $$

(Using Central Limit Theorem)

$$ \begin{aligned} &=1-P(z<1.29) \\ &=1-0.9015 &\left(\begin{array}{l} \text { Using Excel function } \\ \prime \text { "NORMSDIST(1.29)" } \end{array}\right) \\ =0.0985 \end{aligned} $$

From the above results we say that it is more likely to randomly select one student with an \(1 Q\) score greater than \(105,\) because this event having larger probability.