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**1. **QUESTION 1 – Binomial Distribution

**2. **Solution to Question 1 (a) µ = np = 2600 x 0.077 = 200.2
s˛ = npq= 2600 x 0.077 x 0.923 = 184.7
s =13.59
(b) µ - 2s < x < µ + 2s
200.2 – 2 x 13.59 < x < 200.2 + 2 x 13.59
173.02 < x < 227.38
Since 175 is within the range, it is not unusual.

**3. **Question 2 – Binomial Distribution
In a clinical trial of Lipitor, a common drug used to lower cholesterol, 863 patients were given a treatment of 10 – mg Lipitor tablets. That group consists of 19 patients who experienced flu symptoms (based on data from Pfizer, Inc) The probability of the flu symptoms for a person not receiving any treatment is 0.019.
(a) Assuming that Lipitor has no effect on flu symptoms, find the mean and standard deviation for the numbers of people in groups of 863 that can be expected to have flu symptoms.
(b) Based on the result from part (a), is it unusual to find that among 863 people, there are 19 who experience flu symptoms? Why or why not?
(c) Base on the preceding results, do flu symptoms appear to be an adverse reaction that should be concern to those who use Lipitor?

**4. **Solution to Question 2 - Binomial n = 863 p = 0.019
(a) Mean µ = np = 863 x 0.019 = 16.4
Variance s˛ = npq = 863 x 0.019 x 9,981 = 16.08
Standard deviation s = 4
(b) For any x to be usual, we have
µ - 2s < x < µ + 2s
8.4 < x < 24.4
It is not unusual because 19 is within 8.4 and 24.4.
(c) No, because the % is about 1.9 %, not even 2 %.

**5. **Question 3 – Binomial Distribution In a test of the MicroSort method of gender selection, 325 babies are born to couples trying to have baby girls, and 295 of those babies are girls (based on data from the Genetics & IVF Institute).
(a) If the gender selection method has no effect and boys and girls are equally likely, find the mean and standard deviation for the numbers of girls born in groups of 325.
(b) Is the result of 295 girls unusual? Does it suggest that the gender selection method appears to be effective?

**6. **Solution to Question 3 - Binomial (a) µ = np = 325 x 0.5 = 162.5
s˛ = npq = 325 x 0.5 x 0.5 = 81.25
s = 9.01
(b) µ - 2s < x < µ + 2s
162.5 – 2 x 9.01 < x < 162.5 + 2 x 9.01
144.48 < x < 180.52
Since 295 is well outside the range, the method is effective for having girls.

**7. **Question 4 – Poisson Distribution In analyzing hits by V-1 buzz bombs in World War II, South London was subdivide into 576 regions, each with an area of 0.25 km˛ . A total of 535 bombs hit the combined area of 576 regions.
(a) If a region is randomly selected, find the probability that it was hit exactly twice.
(b) Based on the probability found in part(a), how many of the 576 regions are expected to be hit exactly twice?

**8. **Solution to Question 4 - Poisson

**9. ** Question 5 – Poisson Distribution Radioactive atoms are unstable because hey have too much energy. When they release their extra energy, they are said to decay. When studying cesium-137, it is found that during the course of decay over 365 days, 1,000,000 radioactive atoms arereduced to 977,287 radioactive atoms.(a) Find the mean number of radioactive atoms lost through decay in a day. (b) Find the probability that on a given day, 50 radioactive atoms decayed.

**10. **Solution to Question 5 – Poisson

**11. **Question 6 – Poisson Distribution

**12. **Solution to Question 6 - Poisson