Question

Membership in Mensa requires an IQ score above 131.5. Nine candidates take an IQ test, and they have a mean IQ score of 133. IQ scores are normally distributed with a mean of 100 and a standard deviation of 15 (this should be used as the baseline for the questions below.) (a) If one person is randomly selected from the general population, what is the probability of getting someone with an IQ of at least 133?

Answer 1

Answer:

a) $P(X\geq 133)=P(\frac{X-\mu}{\sigma}\geq \frac{133-\mu}{\sigma})=P(Z\geq \frac{133-100}{15})=P(Z\geq 2.2)$

And we can find this probability using the complement rule:

$P(Z\geq 2.2)=1- P(z$

b) $P(\bar X\geq 133)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}\geq \frac{133-\mu}{\sigma_{\bar x}})=P(Z\geq \frac{133-100}{5})=P(Z\geq 6.6)$

And we can find this probability using the complement rule:

$P(Z\geq 6.6)=1- P(z$

c) No. The mean can be lower than 131.5 if we find the probability:

$P(\bar X\leq 133)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}\leq \frac{131.5-\mu}{\sigma_{\bar x}})=P(Z\leq \frac{131.5-100}{5})=P(Z\leq 6.3)$

$P(Z\leq 6.3) \approx 1$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

A.    If 1 person is randomly selected from thegeneral population, find the probability of getting someone with anIQ score of at least 133.

Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:

$X \sim N(100,15)$  

Where $\mu=100$ and $\sigma=15$

We are interested on this probability

$P(X\geq 133)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X\geq 133)=P(\frac{X-\mu}{\sigma}\geq \frac{133-\mu}{\sigma})=P(Z\geq \frac{133-100}{15})=P(Z\geq 2.2)$

And we can find this probability using the complement rule:

$P(Z\geq 2.2)=1- P(z$

And the probability is calculated from the normal standard table or with excel.

B.     If 9 people are randomly selected,find the probability that their mean IQ score is at least 133.

For this case since the distribution for the random variable X is normal then the distribution for the sample mean is also normal and given by:

$\bar X = \sim N(\mu= 100 ,\sigma_{\bar x}= \frac{15}{\sqrt{9}}=5)$

The new z score is defined as

$z=\frac{x-\mu}{\sigma_{\bar x}}$

If we apply this formula to our probability we got this:

$P(\bar X\geq 133)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}\geq \frac{133-\mu}{\sigma_{\bar x}})=P(Z\geq \frac{133-100}{5})=P(Z\geq 6.6)$

And we can find this probability using the complement rule:

$P(Z\geq 6.6)=1- P(z$

And the probability is calculated from the normal standard table or with excel.

C.     Although the results are available,the individual IQ test scores have been lost. Can it be concluded that all 9 candidates have IQ scores above 131.5 so that they allare eligible for Mensa membership?

No. The mean can be lower than 131.5 if we find the probability:

$P(\bar X\leq 133)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}\leq \frac{131.5-\mu}{\sigma_{\bar x}})=P(Z\leq \frac{131.5-100}{5})=P(Z\leq 6.3)$

$P(Z\leq 6.3) \approx 1$

Answer 2

Answer:

0.0139

Step-by-step explanation:

We have to find P(X≥133)

IQ scores normally distributed with mean=100 and standard deviation=sd=15

$P(X\geq 133)=P(\frac{x-mean}{sd} \geq \frac{133-100}{15}$

$P(Z\geq 2.2)=P(0$

$P(Z\geq 2.2)=0.5-0.4861$

$P(Z\geq 2.2)=0.0139$

P(X≥133)=0.0139.

So, if one person is selected at random the probability of getting someone with at least 133 IQ score is 1.39%