# Examples of Stats Homework Solutions

## Solved Statistics Questions

Question #1: Scores on the Wechsler Adult Intelligence Scale (WAIS) approximate a normal curve with a mean of 100 and a standard deviation of 15. What proportion of IQ scores are

(a) above Kristen’s 125?

(b) below 82?

(c) within 9 points of the mean?

(d) more than 40 points from the mean?

Solution: (a) We need to compute the following probability:

$\Pr \left( X\ge {125} \right) = \Pr \left( \frac{X-{100}}{15}\ge \frac{{125}-{100}}{15} \right) = \Pr \left( Z\ge 1.6667 \right)$ $= 1-\Pr \left( Z\le 1.6667 \right) = 1-{0.9522} = {0.0478}$

(b) Now, we need to compute the following probability:

$Pr \left( X\le {82} \right) = \Pr \left( \frac{X-{100}}{15}\le \frac{{82}-{100}}{15} \right) = \Pr \left( Z\le -1.2 \right) = {0.1151}$

(c) Now, we need to compute the following probability:

$\Pr \left( {92}\le X\le {108} \right) = \Pr \left( \frac{{92}-{100}}{15}\le \frac{X-{100}}{15}\le \frac{{108}-{100}}{15} \right)$ $= \Pr \left( -0.5333\le Z\le 0.5333 \right) = \Pr \left( Z\le 0.5333 \right)-\Pr \left( Z\le -0.5333 \right)$ $= {0.7031}-{0.2969} = {0.4062}$

(d) Finally, we need to compute the following probability:

$\Pr \left( X\le {60}\text{ or }X\ge {140} \right) = \Pr \left( \frac{X-{100}}{15}\le \frac{{60}-{100}}{15}\text{ or}\frac{X-{100}}{15}\ge \frac{{140}-{100}}{15} \right)$ $=\Pr \left( Z\le -2.6667\text{ or }Z\ge 2.6667 \right)=1-\Pr \left( Z\le 2.6667 \right)+\Pr \left( Z\le -2.6667 \right)$ $= \Pr \left( Z\le -2.6667\text{ or }Z\ge 2.6667 \right) = 1-\Pr \left( Z\le 2.6667 \right)+\Pr \left( Z\le -2.6667 \right)$ $= 1-{0.9962}+{0.0038} = {0.0077}$

Question #2: On the WAIS measure, IQ scores approximately follow a typical curve with a mean of 100 and a standard deviation of 15. What is the IQ score associated with?

(a) the upper 2 percent, that is, 2 percent to the right (and 98 percent to the left)?

(b) the lower 10 percent?

(c) the upper 60 percent?

(d) the middle 95 percent? [Remember, the middle 95 percent straddles the line perpendicular to the mean (or the 50th percentile), with half of 95 percent, or 47.5 percent, above this line and the remaining 47.5 percent below this line.]

(e) the middle 99 percent?

Solution: (a) We need to compute the following:

$U=\mu +{{\Phi }^{-1}}\left( 1-\alpha \right)\times \sigma ={100}+{{\Phi }^{-1}}\left( 0.98 \right)\times {2}={100}{+} {2.0537}\times {2}={104.1075}$

(b) Now, we need to compute the following:

$L=\mu +{{\Phi }^{-1}}\left( \alpha \right)\times \sigma ={100}+{{\Phi }^{-1}}\left( 0.10 \right)\times {2} = {100}{-} {1.2816}\times {2}={97.4369}$

(c) Now, we need to compute the following:

$U=\mu +{{\Phi }^{-1}}\left( 1-\alpha \right)\times \sigma ={100}+{{\Phi }^{-1}}\left( 0.4 \right)\times {2}={100}{-} {0.2533}\times {2}={99.4933}$

(d) Now, we need to compute the following:

$U= \mu +{{\Phi }^{-1}}\left( 1-\frac{\alpha }{2} \right)\times \sigma ={100}+0.0627\times {2}={100.1254}$ $L=\mu +{{\Phi }^{-1}}\left( \frac{\alpha }{2} \right)\times \sigma ={100}-0.0627\times {2}={99.8746}$

(e) Finally, we need to compute the following:

$U= \mu +{{\Phi }^{-1}}\left( 1-\frac{\alpha }{2} \right)\times \sigma ={100}+0.0125\times {2}={100.0251}$ $L=\mu +{{\Phi }^{-1}}\left( \frac{\alpha }{2} \right)\times \sigma ={100}-0.0125\times {2}={99.9749}$

Question #3: The BMI is measured by dividing a person’s weight (in pounds) by the square of their height (in inches) and multiplying by a multiplier of 703. A BMI of less than 18.5 is considered underweight; 18.5 to 24.9 is considered normal; 25 to 29.9 is considered overweight; and 30 or more is considered obese. It is commonly accepted that Americans have gained weight over the past half-century. Assume that the positively skewed distribution of BMIs for adult American males has a mean of 28 with a standard deviation of 4.

(a) Would the median BMI score exceed, equal, or be exceeded by the mean BMI score of 28?

(b) What z score defines overweight?

(c) What z score defines obese?

Solution: (a) Since it is assumed that the distribution of BMIs for adult American males is positively skewed, it is concluded that the median BMI score will be exceeded by the mean BMI score of 28.

(b) The limits for being overweight are

\begin{align} & {{z}_{1}}=\frac{25-28}{4}=-0.\text{75} \\ & {{z}_{2}}=\frac{29.9-28}{4}=0.\text{475} \\ \end{align}

(c) The z-score that defines obese is

$z=\frac{30-28}{4}=0.5$

Question #4: Television channels sometimes challenge audiences for their views on a broadcast case. Following a televised debate between Barack Obama and Mitt Romney during the 2012 presidential election race in the United States, a news channel sponsored a telephone poll to decide the “winner.” Callers were given two phone numbers, one for Obama and one for Romney, to instantly record their votes.

(a) Comment on whether or not this was a random sample.

(b) How might this poll have been improved?

Solution: (a) This sample is not random, because individuals volunteer to vote, which makes this sampling scheme not random.

(b) The problem with the scheme proposed is that it has a strong likelihood of having voluntary response bias, where those in the population with stronger opinions and stronger desire to express their opinions will be over-represented. One way of improving this problem is to select a real random sample instead.

Question #5: A traditional test for extra-sensory perception (ESP) involves a set of playing cards, each of which shows a different symbol (circle, square, cross, star, or wavy lines). If C represents a correct guess and I an incorrect guess, what is the probability of

(a) C?

(b) CI (in that order) for two guesses?

(c) CCC for three guesses?

(d) III for three guesses?

Solution: (a) Assuming the card is chosen at random, p(C) = 1/5 = 0.2.

(b) Assuming that the cards are chosen with replacement, and assuming independence, we get that p(CI) = 0.2*0.8 = 0.16.

(c) Again, Assuming that the cards are chosen with replacement, and assuming independence, we get that p(CCC) = 0.2*0.2*0.2 = 0.008.

(d) Finally, under the same assumptions as above, p(III) = 0.8*0.8*0.8 = 0.512.

Question #6: A sensor is used to monitor the performance of a nuclear reactor. The sensor accurately reflects the state of the reactor with a probability of .97. But with a probability of .02, it gives a false alarm (by reporting excessive radiation even though the reactor is performing normally), and with a probability of .01, it misses excessive radiation (by failing to report excessive radiation even though the reactor is performing abnormally).

(a) What is the probability that a sensor will give an incorrect report, that is, either a false alarm or a miss?

(b) To reduce costly shutdowns caused by false alarms, management introduces a second completely independent sensor, and the reactor is shut down only when both sensors report excessive radiation. (According to this perspective, solitary reports of excessive radiation should be viewed as false alarms and ignored, since both sensors provide accurate information much of the time.) What is the new probability that the reactor will be shut down because of simultaneous false alarms by both the first and second sensors?

(c) Being more concerned about failures to detect excessive radiation, someone who lives near the nuclear reactor proposes an entirely different strategy: Shut down the reactor whenever either sensor reports excessive radiation. (According to this point of view, even a solitary report of excessive radiation should trigger a shutdown, since a failure to detect excessive radiation is potentially catastrophic.) If this policy were adopted, what is the new probability that excessive radiation will be missed simultaneously by both the first and second sensors?

Solution: (a) The probability that a sensor will give an incorrect report is 0.01 + 0.02 = 0.03

(b) The new probability that the reactor will be shut down because of simultaneous false alarms by both the first and second sensors is p = 0.02*0.02 = 0.0004, assuming that both sensors are independent.

(c) The new probability that excessive radiation will be missed simultaneously by both the first and second sensors is p = 0.01*0.01 = 0.0001, assuming that both sensors are independent.