A manufacturer receives a shipment of 100 parts from a vendor.
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A manufacturer receives a shipment of 100 parts from a vendor.
#The shipment will be unacceptable if more than five of the parts are defective.
#The manufacturer is going to randomly select K parts from the shipment for inspection,
#and the shipment will be accepted if no defective parts are found in the sample.
#How large does K have to be to ensure that the probability that the manufacturer
#accepts an unacceptable shipment is less than 0.1?
#Hint: Use R to plug in different values of K.
# k <- unknown
m <- 6
n <- 94
x <- 0
k <- 0
prob <- dhyper(x, m, n, k)
while (prob > 0.1){
k <- k +1
prob <- dhyper(x, m, n, k)
}
k
## 32
Question 2: (51)
Now suppose that the manufacturer decides to
#accept the shipment if there is at most one defective part in the sample.
#How large does K have to be to ensure that the probability that
#the manufacturer accepts an unacceptable shipment is less than 0.1?
m <- 6
n <- 94
x <- 1
k <- 1
prob <- phyper(x, m, n, k)
while (prob > 0.1){
k <- k + 1
prob <- phyper(x, m, n, k)
}
k
## 51
Question 3: (n)
Answer Let X be the number of trials needed to open the door. Let ‘S 0 denote success i.e. the door is opened and ‘F 0 denote failure for each trial. (a) The event X = x is equivalent to the event F F F . . . F | {z } x−1 times S Also P(S) = 1 n and P(F) = n − 1 n . X has the geometric distribution with p = P(S) = 1 n . Therefore E(X) = 1 p = 1 1 n = n
Question 4: (7)
CHOCOLATE CHIPS IN A COOKIE
#Let the number of chocolate chips in a certain type of cookie have a
#Poisson distribution. We want the probability that a randomly chosen
#cookie has at least two chocolate chips to be greater than 0.99.
#For which of the following values of the mean of the distribution
#is this condition assured? (Please select all that apply!)
# Number of choco chips X ~ Pois(lambda)
# Pr(X > = 2) > 0.99
lambda = 0
prob = 1 - ppois(1, lambda)
while (prob <= 0.99){
lambda = lambda + 0.001
prob = 1 - ppois(1, lambda)
}
lambda
## 6.639
prob
## 0.9900056
Question 5: (400)
Question 6: (E(300 + 10M – 100B + 50*1/4)
Question 7: (342.5)
Workings for both q5. & q6 & q7.
file:///C:/Users/F/AppData/Local/Temp/ksohtml16876/wps1.png
Question 8: (2.5)
Question 9: (0.0833)
Random sample of size 100 is used..
V(2*sample mean) = 4* V(sample mean)/n {since random sample is taken}
Variance of estimator put forward by Researcher A is,
= 4 * (25/12)/100 = 4 * 25/(12*100) = 0.08333
Question 10: (Researcher A)
Question 11: (Theta B)
Question 12:(0.008333)
Random sample of size 1000 is used..(Variance comes down when sample size goes up)
V(2*sample mean) = 4* V(sample mean)/n {since random sample is taken}
Variance of estimator put forward by Researcher A is,
= 4 * (25/12)/1000 = 4 * 25/(12*1000) = 0.008333
Question 13: (n=1000)
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