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Section # 4 Problem # 1 Find the area under the standard normal curve, where Z is between z = -0.43 and z = 0.78 Answer P(-0.43 < z < 0.78) = P(z < 0.78) – P(z < -0.43) = 0.7823 – 0.3336 = 0.4487 Problem # 2 Suppose you must establish regulations concerning the maximum number of people who can occupy a lift. You know that the total weight of 8 people chosen at random follows a normal distribution with a mean of 550kg and a standard deviation of 150kg. What’s the probability that the total weight of 8 people exceeds 600kg? Answer The mean is 550kg and we are interested in the area that is greater than 600kg. z = ( x - μ ) / σ Here x = 600kg, μ = 550kg, σ = 150kg P(x > 600) = P ( z > ( 600 - 550 ) / 150) = 1 – 0.6923

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Section # 4

Problem # 1

Find the area under the standard normal curve, where Z is between z = -0.43 and z = 0.78

Answer

P(-0.43 < z < 0.78) = P(z < 0.78) – P(z < -0.43) = 0.7823 – 0.3336

= 0.4487

Problem # 2

Suppose you must establish regulations concerning the maximum number of people who can occupy a lift. You know that the total weight of 8 people chosen at random follows a normal distribution with a mean of 550kg and a standard deviation of 150kg.

What’s the probability that the total weight of 8 people exceeds 600kg?

Answer

The mean is 550kg and we are interested in the area that is greater than 600kg.

z = ( x - μ ) / σ

Here x = 600kg, μ = 550kg, σ = 150kg

P(x > 600) = P ( z > ( 600 - 550 ) / 150)

= 1 – 0.6923 = 0.3707

Problem # 3

Molly earned a score of 940 on a national achievement test. The mean test score was 850 with a standard deviation of 100. What proportion of students had a higher score than Molly? (Assume that test scores are normally distributed.)

(A)0.10 (B)0.18 (C)0.50 (D)0.82 (E) 0.90

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Answer

The correct answer is B. As part of the solution to this problem, we assume that test scores are normally distributed.

z = (X - μ) / σ

= (940 - 850) / 100

= 0.90

P(Z < 0.90) = 0.8159.

Therefore, the P(Z > 0.90) = 1 - P(Z < 0.90) = 1 - 0.8159 = 0.1841.

Thus, we estimate that 18.41 percent of the students tested had a higher score than Molly.

Problem # 4

A company pays its employees an average wage of $3.25 an hour with a standard deviation of 60 cents. If the wages are approximately normally distributed, determine the proportion of the workers getting wages between $2.75 and $3.69 an hour

Answer

P (2.75 < x < 3.69) = P ([2.75 – 3.25]/0.6 < z < [3.69 – 3.25]/0.6) = P (-0.83 < z < 0.73) = P (z < 0.73) – P (z < -0.83) = 0.7673 – 0.2033 = 0.5640

Problem # 5:

The average life of a certain type of motor is 10 years, with a standard deviation of 2 years. If the manufacturer is willing to replace only 3% of the motors that fail, how long a guarantee should he offer? Assume that the lives of the motors follow a normal distribution.

Answer

P (X < x) = P (Z < k) = 0.03

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From the table:k = -1.88

Therefore: Z = (x - µ) / σ -1.88 = (x – 10)/2

X = -1.88*2+10 = 6.24 years