Suppose that the weights of airline passenger bags are normally distributed with

Question

Suppose that the weights of airline passenger bags are normally distributed with a mean of 49.02 pounds and a standard deviation of 3.83 pounds.

a) What is the probability that the weight of a bag will be less than the maximum allowable weight of 50 pounds? Give your answer to four decimal places.  

b) Let X represent the weight of a randomly selected bag. For what value of c is P(E(X) - c < X < E(X) + c)=0.82? Give your answer to four decimal places.  

c) Assume the weights of individual bags are independent. What is the expected number of bags out of a sample of 17 that weigh less than 50 lbs? Give your answer to four decimal places.  

d) Assuming the weights of individual bags are independent, what is the probability that 11 or fewer bags weigh less than 50 pounds in a sample of size 17? Give your answer to four decimal places.

Explanation / Answer

(a) z = (50-49.02)/3.83 = 0.25

P(X < 50) = P(z<0.25) = 0.5987

(b) Let X represent the weight of a bag. Since the normal pdf is symmetric about E(X), we are looking for the c such that P(E(X) - c < X) = P(X > E(X) + c) = (1-.82)/2 = 0.09. Using the StatCrunch Normal calculator under the Stat > Calculators menu with a mean of 49.02 and a standard deviation of 3.83, we find the area to the left of 42.616453 is equal to 0.06. We then have 42.616453 = 49.02 - c which solving for c yields c = 6.3936

(c) Expected value = 17*0.5987 = 10.1779

(d) P(x<=8) = 1 – P(x=12) – P(x = 13) – P(x=14) – P(x=15) – P(x=16)

= 1 – C(17,12)*0.5987^12*(1-0.5987)^5 – C(17,13)*0.5987^13*(1-0.5987)^4 – C(17,14)*0.5987^14*(1- 0.5987)^3– C(17,15)*0.5987^15*(1-0.5987)^2– C(17,16)*0.5987^16*(1-0.5987)^1

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