5. The US Energy Department states that 60% of all US households have ceiling fa

Question

5. The US Energy Department states that 60% of all US households have ceiling fans. In addition, 29% of all US households have an outdoor grill. Suppose 13% of all US households have both a ceiling fan and an outdoor grill. If a US household is randomly selected, the probability that the household has neither a ceiling fan nor an outdoor grill is: (Hint: same as question 1) b. 0.47 c. 0.76 d. 0.24 6. How does distribution (polygon) A differ from distribution (polygon) B? a. different means, same dispersion (variances) b. Different dispersions (variances), same mean. c. different means and different dispersions (variances) d. None of the above 7. The mpg (miles per gallon) for a certain make of a car is normally distributed with a mean (H) of 31 mpg and a standard deviation () of 0.8 mpg. What is the probability that the mpg for a randomly selected car of this make to be less than or equal to 32 mpg? a. 0.3944 b. 0.8944 c. 0.1056 d. 0.5596 8. If number of trial (n)-20 and probability of success at each trial (p)-0.15,then PX-4)- a. 0.005 b. 0.1821 c. 0.2428 d. 0.8298 9. When probability of success (p)-0.5, the binomial distribution will be a. symmetric b. symmetric if n (sample size) is large c. highly right-skewed d. highly left-skewed

Explanation / Answer

5) P(ceiling fan) = 0.6

P(outdoor grill) = 0.29

P(both) = 0.13

P(ceiling fan or outdoor grill) = 0.6 + 0.29 - 0.13 = 0.76

P(neither ceiling fan not outdo grill) = 1-0.76 = 0.24

Option-D) 0.24

6) Option-B)

7) P(X < 32) = P(Z < (32 - 31) / 0.8)

= P(Z < 1.25)

= 0.8944

Option-B) 0.8944

8) P(X = 4) = 20C4 * 0.154 * (1-0.15)16 = 0.1821

Option-B) 0.1821

9) Option-A)

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