Question

1)The breaking strengths of nylon fibers in dynes are normally distributed with a mean of 12058...

1)The breaking strengths of nylon fibers in dynes are normally distributed with a mean of 12058 and a variance of 200043.

What is the probability that a fiber strength is less than 12550?

Round your answers to the nearest thousandth.

2) The breaking strengths of nylon fibers in dynes are normally distributed with a mean of 12313 and a variance of 200610.

What is the probability that a fiber strength is between12453 and 13977?

Round your answers to the nearest thousandth.

3) The breaking strengths of nylon fibers in dynes are normally distributed with a mean of 11951 and a variance of 200118.

What is the probability that a fiber strength is more than 12450?

Round your answers to the nearest thousandth.

4) A new 1.5-volt battery has an actual voltage that is uniformly distributed between 1.43 and 1.79 volts. Estimate the probability that the sum of the voltages from 119 new batteries lies between 179 and 189 volts.

Round your answers to the nearest thousandth.

5)Suppose that X ∼ N(-1.9,2.7), Y ∼ N(3.5,1.4), and Z ∼ N(1.2, 1.0) are independent random variables. Find the probability that 2.2X + 3Y + 4Z ≥ 8.8.

Round your answer to the nearest thousandth.

6) Suppose that X ∼ N(-2.0,2.6), Y ∼ N(3.0,2.0), and Z ∼ N(1.7, 0.5) are independent random variables.

Find the probability that |3.1X + 3Y + 4Z| ≥ 8.0.

Round your answer to the nearest thousandth.

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Answer #1

1)

Here we have

μ 120581 σ2-200043

The z-score for X = 12550 is

x-μ 12550-12058 _ơ V200043 := 1.10

The probability that a fiber strength is less than 12550 is

P(X < 12550)-P(z < 1.10) = 0.8643

Answer: 0.864

2)

Here we have

μ-12313.0 2006 10

The z-score for X = 12453 is

r_μ 12453-12313 σ V200610 = 0.31

The z-score for X = 13977 is

13977-12313 V2006 10 x-11 3.72 ơ

The probability that a fiber strength is between 12453 and 13977

P( 12453 < X < 13977) P(0.31 < < 3.72) 0.3782

Answer: 0.378

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