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. Incorrect: Your answer is incorrect.

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

NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 49.02
standard Deviation ( sd )= 3.83
a.
P(X < 50) = (50-49.02)/3.83
= 0.98/3.83= 0.2559
= P ( Z <0.2559) From Standard Normal Table
= 0.601
b.
LOWER THAN
P ( Z < x ) = 0.09
Value of z to the cumulative probability of 0.09 from normal table is -1.3408
P( x-u/s.d < x - 49.02/3.83 ) = 0.09
That is, ( x - 49.02/3.83 ) = -1.3408
--> x = -1.3408 * 3.83 + 49.02 = 43.8849
GREATER THAN
P ( Z > x ) = 0.09
Value of z to the cumulative probability of 0.09 from normal table is 1.3408
P( x-u / (s.d) > x - 49.02/3.83) = 0.09
That is, ( x - 49.02/3.83) = 1.3408
--> x = 1.3408 * 3.83+49.02 = 54.1551
P(E(X) - c < X < E(X) + c)=0.82
43.8849 < X < 54.1551 = 0.82
49.02-5.1351 < X < 49.02+5.1351
value of c = 5.1351
c.
P(X < 50) = (50-49.02)/3.83/ Sqrt ( 17 )
= 0.98/0.9289= 1.055
= P ( Z <1.055) From Standard NOrmal Table
= 0.8543
d.
probability of less than 50 lbs is = 0.601
P( X < = 11) = P(X=11) + P(X=10) + P(X=9) + P(X=8) + P(X=7) + P(X=6) + P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)
= ( 17 11 ) * 0.601^11 * ( 1- 0.601 ) ^6 + ( 17 10 ) * 0.601^10 * ( 1- 0.601 ) ^7 + ( 17 9 ) * 0.601^9 * ( 1- 0.601 ) ^8 + ( 17 8 ) * 0.601^8 * ( 1- 0.601 ) ^9 + ( 17 7 ) * 0.601^7 * ( 1- 0.601 ) ^10 + ( 17 6 ) * 0.601^6 * ( 1- 0.601 ) ^11 + ( 17 5 ) * 0.601^5 * ( 1- 0.601 ) ^12 + ( 17 4 ) * 0.601^4 * ( 1- 0.601 ) ^13 + ( 17 3 ) * 0.601^3 * ( 1- 0.601 ) ^14 + ( 17 2 ) * 0.601^2 * ( 1- 0.601 ) ^15 + ( 17 1 ) * 0.601^1 * ( 1- 0.601 ) ^16 + ( 17 0 ) * 0.601^0 * ( 1- 0.601 ) ^17
= 0.7333

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