and
.Chapter Four
Basic probability facts.
Know how to compute the mean, the variance, and the standard deviation of a discrete random variable.
Know the definitions of and.
Know when the binomial distribution applies and how to use it.
Know how to use the formulas
Know what the central limit theorem is.
Know how to compute a confidence interval.
Know how to formulate a null hypothesis, an alternative hypothesis.
Know how to do a significance test.
1. The original simple form of the Connecticut state lottery awarded the following prizes for each 100,000 tickets sold. The winners were chosen by drawing tickets at random.
| Prize Amount | Number Awarded |
|---|---|
| 1 | $5000 |
| 18 | $200 |
| 120 | $25 |
| 270 | $20 |
If you hold one ticket in this lottery, what is your probability of winning anything? What is the mean amount of your winnings?
Solution: The probability of winning anything is
The mean amount of winnings is
2. People with type O-negative blood are universal donors. That is, any patient can receive a transfusion of O-negative blood. Only 7% of the American population have O-negative blood. If 10 people appear at random to give blood, what is the probability that at least one of them is a universal donor?
Solution: The probability that at least one is a universal donor can be computed by using the complement rule.
Since the probability of being a universal donor is 0.07, the probability of being non-universal is 0.93. Thus
3. According to a market research firm, 52% of all residential telephone numbers in Los Angeles are unlisted. A telemarketing company uses random digit dialing equipment that dials residential numbers at random, regardless of whether they are listed in the telephone directory. The firm calls 500 numbers in Los Angeles. What is the exact distribution of the number X of unlisted numbers that are called? Use a suitable approximation to calculate the probability that at least half the numbers called are unlisted.
Solution: Since there are only two outcomes and the 500 numbers are selected independently of each other, the binomial distribution applies in this situation. The exact distribution for the number of unlisted numbers called is B(500,0.52)(k). Since n is fairly large, we can approximate the binomial distribution with a normal distribution where
The probability that at least half of the numbers called are unlisted is
To compute this probability from the normal distribution, we have to convert X = 250 to a z-score.
According to the normal curve table, approximately 0.185 of the curve is below this z. Thus about 81.5% of the curve is above that z, and we have an 81.5% chance of at least half the calls being to unlisted numbers.
4. Patients with chronic kidney failure may be treated by dialysis, using a machine that removes toxic wastes from the blood, a function normally performed by the kidneys. Kidney failure and dialysis can cause other changes, such as retention of phosphorus, that must be corrected by changes in diet. A study of the nutrition of dialysis patients measured the level of phosphorus in the blood of several patients on six occasions. Here are the data for one patient (in milligrams of phosphorus per deciliter of blood):
5.6 5.1 4.6 4.8 5.7 6.4
The measurements are separated in time and can be considered to be an SRS of the patient's blood phosphorus level. Assuming that this level varies normally with 
mg/dl, give a 90% confidence interval for the mean blood phosphorus level. The normal range of phosphorus in the blood is considered to be 2.6 to 4.8 mg/dl. Is there strong evidence that the patient has a mean phosphorus level that exceeds 4.8?
Solution: The six values reported here have a mean value of
The confidence interval is
where z* is chosen so that 90% of the normal curve lies between -z* and z*. Table D on page T-11 lists values of z* for various confidence levels C, and says that for C = 90%, z* = 1.645. Thus the 90% confidence interval is
If we formulate an alternative hypothesis that the patient's 
is greater than 4.8, we can compute a significance level by asking "if 
, what is the probability that we would measure 
by pure chance?" To compute this probability, we compute a z-score for the observed mean.
According to the normal probability table, 1 - 0.9382 = 0.0618 of the curve lies above this z. This is not significant at the usual 5% level.