Question

6.5 - 6) 7) and 8)

Suppose x has a distribution with μ = 11 and σ = 9.

(a) If a random sample of size n = 36 is drawn, find μ_x, σ_x and P(11 ≤ x ≤ 13). (Round σ_x to two decimal places and the probability to four decimal places.)

μx =

σx =

P(11 ≤ x ≤ 13) =

(b) If a random sample of size n = 65 is drawn, find μ_x, σ_x and P(11 ≤ x ≤ 13). (Round σ_x to two decimal places and the probability to four decimal places.)

μx =

σx =

P(11 ≤ x ≤ 13) =

(c) Why should you expect the probability of part (b) to be higher than that of part (a)? (Hint: Consider the standard deviations in parts (a) and (b).)
The standard deviation of part (b) is  ---Select--- smaller than the same as larger than part (a) because of the  ---Select--- larger same smaller sample size. Therefore, the distribution about μ_x is

Question 7 )

Coal is carried from a mine in West Virginia to a power plant in New York in hopper cars on a long train. The automatic hopper car loader is set to put 84 tons of coal into each car. The actual weights of coal loaded into each car are normally distributed, with mean μ = 84 tons and standard deviation σ = 0.7 ton.

(a) What is the probability that one car chosen at random will have less than 83.5 tons of coal? (Round your answer to four decimal places.)
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(b) What is the probability that 39 cars chosen at random will have a mean load weight x of less than 83.5 tons of coal? (Round your answer to four decimal places.)
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Question 8) Suppose the heights of 18-year-old men are approximately normally distributed, with mean 71 inches and standard deviation 2 inches.

(a) What is the probability that an 18-year-old man selected at random is between 70 and 72 inches tall? (Round your answer to four decimal places.)


(b) If a random sample of twenty-five 18-year-old men is selected, what is the probability that the mean height x is between 70 and 72 inches? (Round your answer to four decimal places.)


(c) Compare your answers to parts (a) and (b). Is the probability in part (b) much higher? Why would you expect this?

The probability in part (b) is much higher because the standard deviation is larger for the x distribution.

The probability in part (b) is much higher because the mean is smaller for the x distribution.    

The probability in part (b) is much higher because the mean is larger for the x distribution.

The probability in part (b) is much higher because the standard deviation is smaller for the x distribution.

The probability in part (b) is much lower because the standard deviation is smaller for the x distribution.

Answer 1

To find the probability, we need to find the z scores.

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(Q6) Given = 11 and = 9

(a) n = 36

= = 11

= 9 / sqrt(36) = 9 / 6 = 1.5

P(11 < X < 13) = P(X < 13) - P(X < 11)

For P(X < 13), z = (13 - 11) / 1.5 = 1.33

The probability for P(X < 13) = 0.9082

For P(X < 11), z = (11 - 11) / 1.5 = 0

The probability for P(X < 11) = 0.5

Therefore P(11 < X < 13) = 0.9082 - 0.5 = 0.4082

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(a) n = 65

= = 11

= 9 / sqrt(65) = 1.12

P(11 < X < 13) = P(X < 13) - P(X < 11)

For P(X < 13), z = (13 - 11) / 1.12 = 1.72

The probability for P(X < 13) = 0.9633

For P(X < 11), z = (11 - 11) / 1.5 = 0

The probability for P(X < 11) = 0.5

Therefore P(11 < X < 13) = 0.9633 - 0.5 = 0.4633

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(c) The standard deviation of part (b) is smaller than (a) because of the larger sample size.

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(Q7) Given = 84 and = 0.7

(a) P(X < 83.5)

Z = (83.5 - 84) / 0.7 = -.0.71

Therefore the required probability = 0.2389

(b) n = 39

Z = (83.5 - 84) / [0.7 / sqrt(39)] = -.4.46

Therefore the required probability = 0.0000

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(Q8) Given = 71 and = 2

(a) n = 1

P(70 < X < 72). Since the 2 values are equally distributed about the mean, the required probability is = 2 * P(X < 72) - 1

For P(X < 72), z = (72 - 71) / 2 = 0.5

The probability for P(X < 72) = 0.6915

Therefore the required probability = (2 * 0.6915) - 1 = 0.3830

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(b) n = 25

P(70 < X < 72). Since the 2 values are equally distributed about the mean, the required probability is = 2 * P(X < 72) - 1

For P(X < 72), z = (72 - 71) / [2 / sqrt(25)] = 2.5

The probability for P(X < 72) = 0.9938

Therefore the required probability = (2 * 0.9938) - 1 = 0.9876

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(c) Option 4: The probability in (b) is much higher because the standard deviation is smaller for the x distribution.

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