Download 8. Gallup Poll: A Gallup poll of 1236 adults showed that 14% believe

8. Gallup Poll: A Gallup poll of 1236 adults showed that 14% believe that bad luck follows if your path is
crossed by a black cat, so n = 1236 and p = 0.14. (Assume that a procedure yields a binomial distribution
with n trials and the probability of success for one trial is p. use the given values of n and p to find the
mean μ and standard deviation σ. Also use the range rule of thumb to find the minimum usual value μ 2 σ and the maximum usual value μ + 2 σ)
𝑚𝑒𝑎𝑛 = 𝜇 = 𝑛𝑝 = 1236 × 0.14 = 173.04
Standard deviation = 𝜎 = √𝑛𝑝(1 − 𝑝) = √1236 × 0.14 × 0.86 = √148.8144 = 12.20
Minimum usual value = 𝜇 − 2𝜎 = 173.04 − 2 × 12.20 = 148.64
Maximum usual value = 𝜇 + 2𝜎 = 173.04 + 2 × 12.20 = 197.44
12. Are 24% of M&M’s Blue? Mars, Inc., claims that 24% of its M&M plain candies are blue. A sample of
100 M&M’s is randomly selected.
a. Find the mean and standard deviation for the numbers of blue M&Ms in such groups of 100.
Mean = 𝜇 = 𝑛𝑝 = 100 × 0.24 = 24
Standard deviation = 𝜎 = √𝑛𝑝(1 − 𝑝) = √100 × 0.24 × 0.76 = √18.24 = 4.27
b. Data Set 18 in Appendix B consists of a random sample of 100 M&Ms in which 27 are blue. Is this
result unusual? Does it seem that the claimed rate of 24% wrong?
Let 𝑋 be number of blues. 𝑋 follows binomial distribution with parameters 𝑛 = 100 and 𝑝 = 0.24 under
the claim.
𝑋 − 𝜇 26.5 − 24
𝑃(𝑋 ≥ 27) = 𝑃(𝑋 ≥ 26.5) = 𝑃 (
≥
) = 𝑃(𝑍 ≥ 0.5854) = 1 − 𝑃(𝑍 < 0.5854)
𝜎
4.27
= 1 − 0.7208 = 0.2792
The minimum usual value is 𝜇 − 2𝜎 = 24 − 2 × 4.27 = 18.46 ≅ 19
The maximum usual value is 𝜇 + 2𝜎 = 24 + 2 × 4.27 = 32.54 ≅ 33
So the value 27 which lies between the minimum usual value and the maximum usual value is usual
(We have used 26.5 instead of 27 because binomial distribution is discrete and normal distribution is
continuous. Then normal approximation of binomial distribution is used to compute the probabilities.)
Since an event with probability 0.2792 cannot be considered as an unusual event, the claim cannot be
rejected.
18. Cell Phones and Brain Cancer: In a study of 420,095 cell phone users in Denmark, it was found that
135 developed cancer of the brain or nervous system. If we assume that the use of cell phones has no
effect on developing such cancer, then the probability of a person having such a cancer is 0.000340.
a. Assuming that cell phones have no effect on developing cancer, find the mean and standard deviation
for the numbers of people in groups of 420,095 that can be expected to have cancer of the brain or
nervous system.
𝑛 = 420,095
𝑝 = 0.000340
𝑚𝑒𝑎𝑛 = 𝜇 = 𝑛𝑝 = 420,095 × 0.000340 = 142.83
Standard deviation = 𝜎 = √𝑛𝑝(1 − 𝑝) = √420.095 × 0.000340 × 0.99966 = √142.7837 = 11.95
b. Based on the results from part (a), is it unusual to find the amount 420,095 people, there are 135
cases of cancer of the brain or nervous system? Why or why not?
𝜇 − 2𝜎 = 142.83 − 2 × 11.95 = 118.93 ≅ 119
𝜇 + 2𝜎 = 142.83 + 2 × 11.95 = 166.73 ≅ 167
Since the value 135 lies between the minimum usual value 119 and the maximum usual value 167, it is
not unusual.
c. What do these results suggest about the publicized concern that cell phones are a health danger
because they increase the risk of cancer of the brain or nervous system?
The results suggest that the publicized concern is baseless. The data do not provide sufficient evidence
to reject the hypothesis that cell phones do not affect the incidence of brain cancer.
8. If μ = 1/6, find P(0)
(Assume that the Poisson distribution applies, and proceed to use the given mean to find the indicated
probability)
𝑃(0) =
𝑒 −𝜇 𝜇0
= 𝑒 −𝜇 = 𝑒 −1/6 = 0.8465
0!
12. Deaths from Horse Kicks: A classical example of the Poisson distribution involves the number of
deaths caused by horse kicks to men in the Prussian Army between 1875 and 1894. Data for 14 corps
were combined for the 20-year period, and the 280 corps-years included a total of 196 deaths. After
finding the mean number of deaths per corps-year, find the probability that a randomly selected corps-
year has the following numbers of deaths. The actual results consisted of these frequencies: 0 deaths (in
144 corps-years); 1 death (in 91 corps-years); 2 deaths (in 32 corps-years); 3 deaths (in 11 corps-years); 4
deaths (in 2 corps-years). Compare the actual results to those expected by using the Poisson
probabilities. Does the Poisson distribution serve as a good device for predicting the actual results?
Mean = 𝜇 =
0×144+1×91+2×32+3×11+4×2
280
196
= 280 = 0.7
The Poisson probabilities are
𝑃(0) =
𝑒 −0.7 0.70
= 0.4966
0!
𝑃(1) =
𝑒 −0.7 0.71
= 0.3476
1!
𝑃(2) =
𝑒 −0.7 0.72
= 0.1217
2!
𝑃(3) =
𝑒 −0.7 0.73
= 0.0284
3!
𝑒 −0.7 0.74
𝑃(4) =
= 0.0050
4!
Mulltiplying the probabilities by 280, the expected frequencies are
139.04, 97.33, 34.07, 7.95, and 1.39 and the observed frequencies are 144, 91, 32, 11 and 4
Poisson distribution is a good device for predicting actual results.
16. Life Insurance: There is a 0.9986 probability that a randomly selected 30 year old male lives through
the year (based on data from the US Department of health and Human Services). A Fidelity life insurance
company charges $161 for insuring that the male will live through the year. If the male does not survive
the year, the policy pays out $100,000 as a death benefit. Assume that the company sells 1300 such
policies to 30 year old males, so it collects $209,300 in policy payments. The company will make a profit
if the number of deaths in this group is two or fewer.
a. What is the mean number of deaths in such groups of 1300 males?
Mean number of deaths = 𝑛𝑝 = 1300 × (1 − 0.9986) = 1300 × 0.0014 = 1.82
b. Use the Poisson distribution to find the probability that the company makes a profit from the 1300
policies.
𝑃(𝑐𝑜𝑚𝑝𝑎𝑛𝑦 𝑚𝑎𝑘𝑒𝑠 𝑎 𝑝𝑟𝑜𝑓𝑖𝑡) = 𝑃(𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑑𝑒𝑎𝑡ℎ𝑠 ≤ 2) = 𝑃(0) + 𝑃(1) + 𝑃(2)
=
𝑒 −1.82 1.820 𝑒 −1.82 1.821 𝑒 −1.82 1.822
+
+
= 0.1620 + 0.2949 + 0.2683 = 0.7253
0!
1!
2!
c. Use the binomial distribution to find the probability that the company makes a profit from the 1300
policies, and then compare the result to the result found in part (b).
𝑃(𝑐𝑜𝑚𝑝𝑎𝑛𝑦 𝑚𝑎𝑘𝑒𝑠 𝑎 𝑝𝑟𝑜𝑓𝑖𝑡) = 𝑃(𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑑𝑒𝑎𝑡ℎ𝑠 ≤ 2) = 𝑃(0) + 𝑃(1) + 𝑃(2)
1300
1300
=(
) × 0.00140 × 0.99861300 + (
) × 0.00141 × 0.99861299
0
1
1300
+(
) × 0.00142 × 0.99861298 = 0.1618 + 0.2949 + 0.2686 = 0.7253
2