Question

21. In a study conducted for the State Department of Education, 30% of the teachers who left teaching did so because they were laid off. Assume that we randomly select 10 teachers who have recently left their profession. Find the probability that more than 4 of them were laid off A. .6496 B. .8497 C. .2001 D. .1503 Chapter 7 22. The area under the normal curve between Z-0 and Z 1 is under the normal curve between Z-1 and Z 2 A. Less than B. Greater than C. Equal to D. A, B or C above dependent on the value of the mean E. A. B or C above dependent on the value of the standard deviation the area 23. If the random variable X has a mean of u and a standard deviation o, then (X-p)o has a mean and standard deviation respectively: A. and ơ B. X ands C, 0 and 1 D. 1 and 0 24. The fill weight of a certain brand of adult cereal is normally distributed with a mean of 910 grams and a standard deviation of 5 grams. If we select one box of cereal at random from this population, what is the probability that it will weigh less than 904 grams? A. .8849 B. .3849 C. .1151 D. 7698 E. 2302 25. The normal approximation of the binomial distribution is appropriate when: A. np 2 10 B. n(1-p) 2 10 C. np s 10

Answer 1

It is given that 30% of the teachers who left teaching did so because they were laid off. Let the sample size be n = 10. Let X be the number of teachers who left teaching did so because they were laid off. X - Bin(n=10, p=0.3) The probability mass function is given as follows: P(x = x)=) p*(1-p)**; x = 0,1,2...n Compute the probability that more than 4 of them were laid off. That is, P(X >4)=1-P(X 34) 3 P°(1-phº-11 (1-2). (60°) pº{1- p)2-(19) p(1– p) - (40) p*(1- p) = 0.1503 Therefore, the probability that more than 4 of them were laid off is P(X > 4) = 0.1503 Thus, the correct option is D. 0.1503. The area under the normal curve between Z=0 and Z=1 is given as follows: P(0 <Z <1)=P(Z <1)-P(Z <0) = 0.8413 -0.5 (from standard normal tables) =0.3413 The area under the normal curve between Z=1 and Z=2 is given as follows:

P(1<Z<2)=P(Z <2)-P(Z <1) -0.9772 -0.8413 from standard normal tables) = 0.1359 Thus, it can be concluded that the area under the normal curve between Z=0 and Z=1 is greater than the area under the normal curve between Z=1 and Z=2. Therefore, option B. Greater than is the correct option. 23. It is given that a random variable X has a mean y and standard deviation o. The variable is a standardized variable and have mean 0 and standard deviation 1. 0 Therefore. 1 has a mean 0 and standard deviation 1. Thus, the correct option is C. 0 and 1. 24. Let the fill weight of a cereal be X. It is given that the fill weight of a cereal is normally distributed with a mean of 910 grams and a standard deviation of 5 grams. That is, X-Nu=910,0 =5) Compute the probability that it will weigh less than 904 grams. That is, PLX <904) – P{ 4_9045910) = P(Z<-1.2) = 0.1151 Therefore, the correct option is Option C. 0.1151.

25. The normal approximation is not accurate for small values of , thus the normal approximation of the binomial distribution is appropriate when np 210 and n(1-p)210. The number of observations n must be large enough and the value of p should be small. Thus, the correct option is option A np 210 and option B n(1-p210 both should hold. Thus, the normal approximation of the binomial distribution is appropriate when np 2 10 and n(1-P) 10.