Question

1) The number of students taking the Scholastic Aptitude Test (SAT) has risen to an all-time high of more than 1.5 million. Students are allowed to repeat the test in hopes of improving the score that is sent to college and university admission offices. The number of times the SAT was taken and the number of students are as follows. (HINT: This is a Discrete Probability Distribution Problem) Number of Times SAT is taken Number of Students 1 721,769 2 601,325 3 166,736 4 22,299 5 6,730 a. Let x be a random variable indicating the number of times a student takes the SAT. Show the probability distribution for this random variable. b. What is the probability that a student takes the SAT more than one time? c. What is the probability that a student takes the SAT three or more times? d. What is the expected value of the number of times the SAT is taken? What is your interpretation of the expected value? e. What is the variance and standard deviation for the number of times the SAT is taken?

2) In San Francisco, 30% of workers take public transportation daily. In a sample of 10 workers, a. Clearly state what the random variable in this problem is? b. What is an appropriate distribution to be used for this problem and why? c. What is the probability that exactly three workers take public transportation daily? d. What is the probability that NONE of the workers take public transportation daily? e. What is the probability that more than five workers take public transportation daily? f. What is the probability that less than seven workers take public transportation daily? g. What is the probability at least two but no more than eight workers take public transportation daily?

3) In a typical month, an insurance agent presents life insurance plans to 40 potential customers. Historically, one in four such customers chooses to buy life insurance from this agent. You may treat this as a binomial experiment. a. What is the probability of success? b. What is the total number of trials? c. Create a probability distribution table which includes probability of each possible outcome. Also create the cumulative probability column. d. What is the probability that exactly five customers will buy life insurance from this agent in the coming month? e. What is the probability that no more than 10 customers will buy life insurance from this agent in the coming month? f. What is the probability that at least 20 customers will buy life insurance from this agent in the coming month? g. Determine the mean and standard deviation of the number of customers who will buy life insurance from this agent in the coming month? h. What is the probability that the number of customers who buy life insurance from this agent in the coming month will lie within two standard deviations of the mean?

4) During the period of time that a local university takes phone-in registrations, calls come in at the rate of one every two minutes. a. Clearly state what the random variable in this problem is? b. What is an appropriate distribution to be used for this problem and why? c. What is the expected number of calls in one hour? d. What is the probability of receiving three calls in five minutes? e. What is the probability of receiving NO calls in a 10-minute period? f. What is the probability of receiving more than five calls in a 10-minute period? g. What is the probability of receiving less than seven calls in 15-minutes? h. What is the probability of receiving at least three but no more than 10 calls in 12 minutes?

5) The annual number of industrial accidents occurring in a particular manufacturing plant is known to follow a Poisson distribution with mean 12. a. What is the probability of observing of observing exactly 12 accidents during the coming year? b. What is the probability of observing no more than 12 accidents during the coming year? c. What is the probability of observing at least 15 accidents during the coming year? d. What is the probability of observing between 10 and 15 accidents (including 10 and 15) during the coming year? e. Find the smallest integer k such that we can be at least 99% sure that the annual number of accidents occurring will be less than k.

Answer 1

1) SAT PROBLEM

a)

Let X: Number of times the student took SAT

X can take 5 values i.e. x=1,2,3,4 or 5.

The probability distribution of this random variable is calculated in the following table:

Here the third column P(X=x) is calculated by dividing the number of students taking SAT the x^th time with the total number of students. For example: When x=3, P(X=3)= 166736/1518859 =0.109777.

b)

We have to find the P(Student takes SAT more than once) = P(X>1)

Now P(X>1) = P(X=2)+P(X=3)+P(X=4)+P(X=5)

= 0.395906+0.109777+0.014681+0.004431

= 0.524795

c)

We have to find the P(Student takes SAT three or more times) = P(X=3)+P(X=4)+P(X=5)

= 0.109777+0.014681+0.004431

= 0.128889   

d)

The expected number of times SAT was taken is given by the formula:

The summation is taken over 1 to 5.

E(X) = 1*P(X=1)+2*P(X=2)+3*P(X=3)+4*P(X=4)+5*P(X=5)

= 1.677228

The expected number of times the student will take SAT is 1.677228, rounded off to nearest integer to be 2.

e)

Now the formula for variance is

Calculating E(X²) = 1²*P(X=1)+2²*P(X=2)+3²*P(X=3)+4²*P(X=4)+5²*P(X=5)

= 3.392499

Hence

Var(X) = (3.392499) - (1.677228)²

⁼ 0.579404

And Standard Deviation

=(0.579404)^0.5

= 0.761186

2) TRANSPORT PROBLEM

Let p = P(Workers takes public transport everyday) = 0.30

and q = P(Workers do not take public transport everyday) = 1-p = 0.70

a)

The random variable in the problem is:

X : Number of people taking public transport daily

b)

The appropriate distribution to be used for the problem is "Binomial Distribution" because it perfectly describes this particular situation where we only have either the probability of success OR failure i.e. either the worker takes public transport or they do not take public transport.

P(X=x) = nCx * p^x * (1-p)^10-x

i.e.

P(X=x) = 10Cx * (0.30)^x * (0.70)^10-x

c)

We have to find P(X=3)

P(X=3) = 10C3 * (0.30)³ * (0.70)⁷

= 0.266828

d)

We have to find P(X=0)

P(X=0) = 10C0 * (0.30)⁰ * (0.70)¹⁰

= 0.028248

e)

We have to find P(X>5)

P(X>5) = P(X=6)+ P(X=7)+P(X=8)+P(X=9)+P(X=10)

= 10C6 * (0.30)⁶* (0.70)⁴+10C7 * (0.30)⁷ * (0.70)³+10C8 * (0.30)⁸ * (0.70)²+10C9 * (0.30)⁹ * (0.70)¹

+10C10 * (0.30)¹⁰ * (0.70)⁰

= 0.047349

f)

We have to find P(X<7)

= P(X=6)+ P(X=5)+P(X=4)+P(X=3)+P(X=2)+P(X=1)+P(X=0)

=0.9894079

g)

We have find the probability P(1<X<9)

P(1<X<9) = P(X=2)+ P(X=3)+P(X=4)+P(X=5)+P(X=6)+P(X=7)+P(X=8)

=0.850548