Question

2.  The percentage of people in a population with a certain ailment (Ailment A) is  7.3%.

a.  If you select a sample of  10 people from this population, what is the probability that at most two of them will have Ailment A ?  

b.  What is the probability that at least 3 of them would 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 sample of  200, that at least  20 will have Ailment A ?

4.  The accumulated miles between repairs for vehicle engines is 24,000 miles with a standard deviation of  2000 miles. The accumulated miles, which have been recorded over time, follow a normal distribution.

a.  Find the probability that an engine you just received will last longer than  26,000 miles.

b.  Find the probability that the mean accumulated mileage from a sample of  10 engines exceeds  26,000 miles.

c.  Find the 1^st, 2^nd, and 3^rdquartiles for the accumulated miles between repairs.

d.  Now, you are looking at vehicle transmissions.  The historical data for transmission mileages indicates a population mean of  16,000 miles with a standard deviation of 2600 miles.  The mileage for transmissions does not follow a normal distribution. Find the probability that, in a large train shipment of  40 transmissions, the average mileage for this sample will be less than  15,000 miles.

e.  If the average for your transmission sample of  40 falls below the bottom  10%, you are going to declare a stand-down of the workforce to determine what is going wrong.  What is the cutoff number of miles for the bottom 10% of your sample average?

f.  Back to the engines . . .  If a single engine is considered a “failure” if it doesn’t accumulate at least 22,000 miles between repairs, what is the chance that an engine will 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 are placed into vehicles?

Answer 1

2)

it is a binomial probability distribution,because there is fixed number 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 z score 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