2. The percentage of people i
2. The percentage of people in a population with acertain ailment (Ailment A) is 7.3%.
a. If you select a sample of 10 peoplefrom this population, what is the probability that at most two ofthem will have Ailment A ?
b. What is the probability that at least 3 of themwould have this ailment ?
c. If you select a sample of 200 people,what is the probability that less than 10 will have ailment A? Use the normal approximation technique.
d. What is the probability, in your sampleof 200, that at least 20 will have Ailment A?
4. The accumulated miles between repairs for vehicleengines is 24,000 miles with a standard deviationof 2000 miles. The accumulated miles, which have beenrecorded over time, follow a normal distribution.
a. Find the probability that an engine you justreceived will last longer than 26,000 miles.
b. Find the probability that the mean accumulatedmileage from a sample of 10 enginesexceeds 26,000 miles.
c. Find the 1st, 2nd, and3rdquartiles for the accumulated miles betweenrepairs.
d. Now, you are looking at vehicletransmissions. The historical data for transmissionmileages indicates a population mean of 16,000 mileswith a standard deviation of 2600 miles. The mileage fortransmissions does not follow a normal distribution. Find theprobability that, in a large train shipment of 40transmissions, the average mileage for this sample will be lessthan 15,000 miles.
e. If the average for your transmission sampleof 40 falls below the bottom 10%, you aregoing to declare a stand-down of the workforce to determine what isgoing wrong. What is the cutoff number of miles for thebottom 10% of your sample average?
f. Back to the engines . . . If a singleengine is considered a “failure” if it doesn’t accumulate at least22,000 miles between repairs, what is the chance that an enginewill fail to meet its anticipated mileage accumulation?
g. Given the criteria just stated, what would be the“expected number” of failures in the next 1000 engines that areplaced into vehicles?
Answer:
2)
it is a binomial probability distribution,because there is fixednumber of trials,
only two outcomes are there, success and failure
trails are independent of each other
and probability is given by
| P(X=x) = C(n,x)*px*(1-p)(n-x) |
where
Sample size , n = 10
Probability of an event of interest, p =0.073
a)
P(at most 2) = P(x≤2) = P(x=0) + P(x=1) + P(X=2) = 0.9684
b)
P(X≥3) = 1 – P(X<3) = 0.0316
c)
Binomial to normal approximations using continuity factor
Sample size , n = 200
Probability of an event of interest, p =0.073
left tailed
X <10
Mean = np = 14.6
std dev ,σ=√np(1-p)=3.679
P(Xbinomial <10) = P(Xnormal <9.5)
Z=(Xnormal – µ ) / σ = -1.386
=P(Z<-1.386) = 0.0828
[excel formula for probability from zscore is =NORMSDIST(Z) ]
d)
Sample size , n = 200 Probability of an event of interest, p = 0.073 right tailed X ≥ 20 Mean = np = 14.6 std dev ,σ=√np(1-p)= 3.6789 P(X ≥ 20 ) = P(Xnormal ≥ 19.5 ) Z=(Xnormal – µ ) / σ = 1.332 =P(Z ≥ 1.332 ) = 0.0914
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