Math problems.....Can you tell me how you come up with the answers!!!?

Question

1. A company finds that an average of 20% of all new employees resign during the first year. The probability that amoung the next 20 employees, less then 7 will resign during the first year is: a. .421 b. .198 c. .172 d. .067 e. none of the above

2. The average number of pounds of meat a person consumes a year is 218.4lb. Assume that the standard deviation is 25 lb and the distribution is approximately normal. If a sample of 40 people is selected, the probability that the mean of the sample will be at least 224lb. per year is: a. .5871 b. 9222 c. .4222 d. .0778 e. none of the above


3. Let x~n (100, 256). The third quartle for x is: a. 120.48 b. 140 c. 3/4 d. none of the above

4. Let x~n (8, 16). what is the a such that p(x>a)=.975 a. -1.96 b. 1.96 c. -0.16 d. -15.84 e. 0.16

Answer 1

The key to answering number 1 is to recognize that all the given probabilities are much too small to be correct. Technically, the probability that less than 7 employees will resign during the first year can be calculated by using the binomial distribution with p=0.2 and n=20. If the random variable X represents the number of employees who resign during the first year, this calculation amounts to calculating

P(X<7) = P(X=0) + P(X=1) + P(X=2)+ ... + P(X=6).

However, since you know that the expected value of the binomial distribution is:

n*p = 20*0.2 = 4 employees,

you should immediately realize that P(X=4)=0.2, allowing you to immediately rule out b., c., and d. since

P(X<7) must surely be greater or equal than P(X=4) = 0.2.

Next, since the variance of the binomial is:

n*p*(1-p) = 20*0.2*0.8 = 3.2,

the standard deviation is sqrt(3.2)=~1.8. Based on the fact that the binomial can be approximated by the normal, you should realize that P(4-1.8 < X < 4+1.8) should be about 0.66. In other words, about 66% of the time, 2 to 6 employees will resign within the first year, a probability higher than the remaining answer, a. (0.421).

Thus, e. none of the above is the correct answer.

The true probability is actually P(X<7)=0.91. Less than 7 employees will resign more than 90% of the time.

Answer 2

I just do them in my head. Much easier than trying to use formulas.