1. Comparing Z-scores
Two students A and B did two different exams. A did Finance and his grade was 7.9. B did Microeconomics and his grade was 7.6 (both grades on a scale from 0-10). The mean grade of all students who did Finance was 7.4 with standard deviation 3; the mean grade of all students who did Microeconomics was 7.5 with standard deviation 0.5.
Who performed better, A or B? Explain your answer.

The probability that a student gets a grade higher than 7.9 for Finance is:

\displayrule{P(X\geq7.9)=P(Z\geq\frac{7.9-7.4}{3})= P(Z\geq{0.1670)}}

The probability that a student gets a grade higher than 7.6 for Microeconomics is:

\displayrule{P(X\geq7.6)=P(Z\geq\frac{7.6-7.5}{0.5})= P(Z\geq{0.2)}}

Since P(Z\geq0.2)<P(Z\geq0.167) it appears more difficult for B to get a 7.6 than it was for A to get a 7.9. So, although B has a lower grade than A, he performed better.

2. Tulips
A certain type of Amsterdam tulip bulb germinates 90% of the time. A backyard farmer planted 25 tulip bulbs. Answer the following questions.
a. What is the probability that exactly 20 germinate?
b. What is the probability that 20 or more germinate?
c. What is the expected number of bulbs that germinate?

a. Here we have a binomial distribution problem with n=25; p=0.9. Use the binomial distribution formulas in Excel to calculate the probability that exactly 20 bulbs germinate (20 successes). The answer is 0.0646 or 6.5%.
b. P(X\geq{20})=1-P(X\leq19)=1-0.0334=0.9666 or 96.7% by using the cumulative binomial distribution.
c. It is obvious that the expected (average) number of bulbs that germinate equals np=22.5.

3. Incoming faxes
After an analysis of incoming faxes the manager of an accounting firm determined the probability distribution of the number of pages per fax as follows:


a. Compute the probability that the number of pages equals 4?
b. How many percent of the faxes has more than 5 pages?
c. Compute the mean and the variance of the number of pages per fax.

a. \sum{P(X)=1, so P(4)=1-0.70=0.30.
b P(X>5) = P(X=6) + P(X=7)=0.10+0.08=0.18. So the percentage is 18%.
c. \mu=E(X)=\sum{xP(x)}=1\cdot(0.05)+\cdots+7\cdot(0.08)=4.00

4. National railroad
The Dutch National Railroad NS found that 85% of all trains arrive on time.
a. If you randomly select 100 train arrivals, compute the probability that exactly 85 of these trains arrive on time.
b. Compute the probability that 85 trains or more arrive on time.
c. Under certain conditions a binomial distribution may be approximated by a normal distribution. What are the mean and standard deviation of the normal distribution that approximates the binomial distribution described above.
d. Repeat the question under b., now using the normal distribution. Explain the difference.

a. This is a binomial experiment with n=100 and p=0.85 (probability of being on time). If X denotes the number of trains on time, Excel computes P(X=85)=0.1111.
b. We compute P(X\geq85)=1-P(X\leq84)=1-0.4317=0.5683..
c. The mean of this binomial experiment is \mu=np=85 and \sigma=\sqrt{np(1-p)}=3.57 which can be used as the parameters of the approximating normal distribution N(85,3.57) .
d. Taking the correction factor for continuity into account we get as approximation P(X\geq84.5)=1-P(X\leq84.5)=1-0.4443=0.5557.

5. Roulette
In a game of roulette, a steel ball is rolled onto a wheel that contains 18 red, 18 black, and 2 green slots. If the ball is rolled 25 times, find the probability of the following events.
a. The ball falls into the green slots two or more times.
b. The ball does not fall into any green slots.
c. The ball falls into black slots more than 5 and less than 15 times.

a. There are 38 slots in total (18+18+2). The probability that the ball falls in a green slot (success) is p=\frac{2}{36}. So, we have a binomial experiment with the total number of experiments n=25 and p=\frac{2}{36} . Excel only computes the probability P(X\leq{k}. So P(X\geq2)=1-P(X\leq1)=1-0.6163=0.3837. (Naturally, this relatively simple calculation can also be done manually).
b. In this case we have n=25, p=\frac{36}{39},  (ball not in green slots) and we compute the probability that 25 times out of 25 experiments the ball does not fall in the green slots:
c. In this case we have n=25, P(ball in black slot) =\frac{18}{38}. We have to compute:
=0.8564-0.0045=0.8520, where we used Excel.

6. Height Dutch women
The heights of Dutch women are normally distributed with a mean of 170.7 cm and a standard deviation of 6.3 cm.
a. What is the probability that a randomly selected woman is taller than 178 cm?
b. A random sample of four women is selected. What is the probability that the sample mean height is greater than 178 cm.
c. A random sample of 100 women is selected. What is the probability that the sample mean height is greater than 178 cm.

a. Use Excel (Cumulative Distribution Function) to compute P(l>178)=0.1233
b. If \overline{l} is the mean height of 4 women, then P(\overline{l}>178)=P(Z>\frac{178-170.7}{6.3/\sqrt{4}})=P(Z>2.32)=0.0102, which can be computed either manually (table) or by Excel.
c. Now the probability is almost.

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