Question

Lab - Chapter 5 - Normal Distribution Problems

1. The machine that packages 5 lb. bags of sugar is designed to put an average (mean) of 5.1 lbs. with a standard deviation of 0.4 lbs. into the package.

a. What is the probability that a bag of sugar weighs less than 5 lbs.?

b. What is the probability that a bag of sugar weighs more than 5.1 lbs.?

c. What is the probability that a bag weighs between 4.5 and 5.5 lbs.?

d. What is the probability that a bag weighs between 5.4 and 6.6 lbs.?

2. A bag of individually wrapped candy claims to contain 45 pieces. The machine packaging the candy is designed to put an average of 43 pieces in the bag with a standard deviation of 4 pieces.

a. What is the probability that the bag has more than 45 pieces?

b. What is the probability that the bag has more than 51 pieces?

c. What is the probability that the bag contains between 43 and 47 pieces of candy?

d. What is the probability that the bag contains between 47 and 53 pieces of candy?

3. In recent years, the results of a particular college entrance exam showed an average score of 55 with a standard deviation of 4 points.

a. Based on these results, what is the probability that an incoming student scores below a 50?

b. What is the probability that a student scores above a 60?

c. What is the probability that a student scores between a 53 and a 59?

d. What is the probability that a student scores below a 58?

4. A bag of potato chips claims to contain 7 oz. of potato chips. Random sampling determines that the bags contain an average of 7.2 oz. with a standard deviation of 0.081 oz.

a. What is the maximum weight that 20% of the bags contain less than?

b. What is the minimum weight that the heaviest 15% of the bags contain?

c. In what weight interval are the middle 60% of the bags contained?

Answer 1

SOLUTION 1a: Since μ=5.1 and σ=0.4 we have:

P ( X<5 )=P ( X−μ<5−5.1 )=P ((X−μ)/σ<(5−5.1)/0.4)

Since (x−μ)/σ=Z and (5−5.1)/0.4=−0.25 we have:

P (X<5)=P (Z<−0.25)

Use the standard normal table to conclude that:

P (Z<−0.25)=0.4013

SOLUTION 1b: Since μ=5.1 and σ=0.4 we have:

P ( X>5.1 )=P ( X−μ>5.1−5.1 )=P (( X−μ)/σ>(5.1−5.1)/0.4)

Since Z=(x−μ)/σ and (5.1−5.1)/0.4=0 we have:

P ( X>5.1 )=P ( Z>0 )

Use the standard normal table to conclude that:

P (Z>0)=0.5

SOLUTION C: P ( 4.5<X<5.5 )=P ( 4.5−5.1< X−μ<5.5−5.1 )=P ((4.5−5.1)/0.4<(X−μ)/σ<(5.5−5.1)/0.4)

Since Z=(x−μ)/σ , (4.5−5.1)/0.4=−1.5 and (5.5−5.1)/0.4=1 we have:

P ( 4.5<X<5.5 )=P ( −1.5<Z<1 )

Use the standard normal table to conclude that:

P ( −1.5<Z<1 )=0.7745

SOLUTION D: P ( 5.4<X<6.6 )=P ( 5.4−5.1< X−μ<6.6−5.1 )=P ((5.4−5.1)/0.4<(X−μ)/σ<(6.6−5.1)/0.4)

Since Z=(x−μ)/σ , (5.4−5.1)/0.4=0.75 and (6.6−5.1)/0.4=3.75 we have:

P ( 5.4<X<6.6 )=P ( 0.75<Z<3.75 )

Use the standard normal table to conclude that:

P ( 0.75<Z<3.75 )=0.2265

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