Question

1)A sample of size 50 will be drawn from a population with mean 73 and standard deviation 8. Find the 19th percentile of . Group of answer choices 73.8 75.8 75.7 72.0

2)Use technology to solve the following problem: A certain car model has a mean gas mileage of 33 miles per gallon (mpg) with a standard deviation A pizza delivery company buys 48 of these cars. What is the probability that the average mileage of the fleet is between 32.7 and Group of answer choices 0.2442 0.7558 0.4884 0.5116

3)Using technology, use the Central Limit Theorem to find the indicated probability. The sample size is n, the population proportion is p, and the sample proportion is . n = 114, p = 0.51; P( > 0.47) Group of answer choices 0.1965 0.7717 0.8035 0.7563

4)For a particular diamond mine, 81% of the diamonds fail to qualify as "gemstone grade". A random sample of 92 diamonds is analyzed. Find the probability that more than 79% of the sample diamonds fail to qualify as gemstone grade. Group of answer choices 0.3594 0.3121 0.6879 0.6406

5)At a cell phone assembly plant, 79% of the cell phone keypads pass inspection. A random sample of 103 keypads is analyzed. Find the probability that more than 83% of the sample keypads pass inspection. Group of answer choices 0.1587 0.8413 0.8749 0.1251

Answer 1

Solution:-
1) option D. 72.00

Explanation : Given that n = 50, mean = 73, sd = 8
P(X < x) = 0.19, z = -0.8779

X = mean + (Z*(sd/sqrt(n))
= 73 - (0.87789*(8/sqrt(50))
= 72.00

3. option C. 0.8035

Explanation: n = 114, p = 0.51; P( > 0.47)

P(P^ > 0.47) = P((p^-p)/sqrt(pq/n) > (0.47-0.51)/sqrt(0.51*0.49/114)
= P(Z > -0.8543)
= 0.8023

4) option C. 0.6879

Explanation : p = 0.81, n = 92, p^ = 0.79

P(p^ > 0.79) = P(Z > (0.79-0.81)/sqrt(0.81*0.19/92)
= P(Z > -0.4889)
= 0.6879

5) option A.

Explanation: P(Z > (0.83-0.79)/(0.79*0.21/103))
P(Z > 0.9967)
= 0.1587