In a population of all NIU students, the proportion of students who live in the

Question

In a population of all NIU students, the proportion of students who live in the dorms is 72%. Suppose you are to randomly select 220 students, and compute the sample proportion (P). Use this information to answer the next two questions. You are interested in the distribution of the sample proportion (P). Assuming a Normal Distribution, what is the mean and standard deviation for the sampling distribution of P? 1. Mean = 0.72, SD a. b. 0.449 Mean= 0.72, SD=0.0009 Mean= 0.72, SD=0.72 C. d, Mean=0.72,SD=0.0303 2, what is the probability that more than 80% of the 200 randomly selected students will live in the doms 0.9959 a. 0.9959 b. 2.64 c. 0.080 d. 0.0041 Suppose we have a population that is multimodal. Under the Central Limit Theorem (CLT), which of the following statements about the sampling distribution of the sample mean, y, is true? 3. a. Regardless of sample size, the sampling distribution of ' is approximately Normal. b. Regardless of sample size, the sampling distribution of is multimodal. c. For large samples, the sampling distribution of is approximately Normal. d. For large samples, the sampling distribution of y is multimodal. The life of a General Electric light bulb, in hours, has a mean of 620, and a standard deviation of 48. Suppose you go to the hardware store and purchase 9 light bulbs, and calculate the mean life (), in hours, of those 9 light bulbs. Use this information to answer the next two questions. 4. The sampling distribution of (sample mean life, in hours, of the 9 sampled light bulbs) is approximately Normal with what mean, and what standard deviation? a. Mean=620. SD=48 b, Mean = 620, SD = 9 C. Mean = 620, SD = 16 d, Mean=620. SD=3 What is the probability that the mean life, in hours, of the 9 sampled light bulbs is less than 600 hours? 5. a. 0.6628 b. 0.8944 c. 1.25 d. 0.1056

Explanation / Answer

1)

mean =p = 0.72

standard deviation = sqrt(p * (1-p)/n) = 0.0303

hence choose d)

2)

P(X > 80%) = P(z > 0.8 -0.72/0.0303)

= P(Z > 2.64)

= 0.0041

3)

c)

4)

mean = 620

SD = 48/sqrt(9)

= 16

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