Question

1. An elevator accident occurred with 16 people inside. The capacity will be exceeded if 16 people have a mean weight greater than 154 pounds. Assume weights of people are normally distributed with a mean of 172.4 pounds and a standard deviation of 25.8 pounds. What is the probability the capacity will be exceeded with 16 random people in the elevator?
Round answer to 4 decimal places.

2. In a study of the health effects of cigarettes, a random sample of 32 filtered cigarettes was obtained and the tar content was measured. The sample has a mean of 19.2 mg. Given that the tar content of cigarettes have a mean of 20.1 mg and a standard deviation of 3.15 mg, what is the probability that 32 filtered cigarettes have a mean of 19.2 mg or less?
Round answer to 4 decimal places.

3.The scores on a test are normally distributed with a mean of 60 and standard deviation of 3. What is the probability that a sample of 55 students will have a mean of at least 60.3162? Round your answer to 4 decimal places.

4.A study of the amount of time it takes a mechanic to rebuild the transmission for a 2005 Chevrolet Cavalier shows that the mean is 8.4 hours and the standard deviation of 1.8 hours. If 38 mechanics are randomly selected, find the probability that their mean rebuild time exceeds 8.7 hours. Round your answer to 4 decimal places.

5.The annual precipitation amounts in a certain mountain range are normally distributed with a mean of 82 inches, and a standard deviation of 9 inches. What is the probability that the mean annual precipitation during 25 randomly picked years will be less than 85.2 inches?

Answer 1

1. Solution :

Given that mean μ = 172.4 , standard deviation σ = 25.8 , n = 16

=> P(x > 154) = P((x - μ)/(σ/sqrt(n)) > (154 - 172.4)/(25.8/sqrt(16)))

= P(Z > -2.8527)

= P(Z < 2.8527)

= 0.9978

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2. Solution :

Given that mean μ = 20.1 , standard deviation σ = 3.15 , n = 32

=> P(x <= 19.2) = P((x - μ)/(σ/sqrt(n)) <= (19.2 - 20.1)/(3.15/sqrt(32)))

= P(Z <= -1.6162)

= 1 - P(Z < 1.6162)

= 1 - 0.9474

= 0.0526

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3. Solution :

Given that mean μ = 60 , standard deviation σ = 3 , n = 55

=> P(x >= 60.3162) = P((x - μ)/(σ/sqrt(n)) >= (60.3162 - 60)/(3/sqrt(55)))

= P(Z >= 0.7817)

= 1 - P(Z < 0.7817)

= 1 - 0.7823

= 0.2177

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4. Solution :

Given that mean μ = 8.4 , standard deviation σ = 1.8 , n = 38

=> P(x > 8.7) = P((x - μ)/(σ/sqrt(n)) > (8.7 - 8.4)/(1.8/sqrt(38)))

= P(Z > 1.0274)

= 1 - P(Z < 1.0274)

= 1 - 0.8485

= 0.1515

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5. Solution :

Given that mean μ = 82 , standard deviation σ = 9 , n = 25

=> P(x < 85.2) = P((x - μ)/(σ/sqrt(n)) > (85.2 - 82)/(9/sqrt(25)))

= P(Z < 1.7778)

= 0.9625