Solution Sample Midterm III.

Numeric computation might not be 100 % correct. Still I thought it is better for you to  have almost correct answer than nothing. Good luck and see you all in spirit on Monday. I will be there at 8:30.

Somehow, web is not printing symbols. Please note the following.

\mu will denote the Greek letter mu and \sigma will denote the Greek letter sigma

<= denote less than or  equal to

<=  denote less than or  equal to

\ge denotes greater than equal to.

>= denotes greater than equal to.

[1] The probability that a person loses weight from a certain diet program is 0.7. Among 40 people in the program,

(a)      What is the probability that at least 30 people lose weight?

(b)      What is the probability that between 16 and 18 people lose weight?

(c)      Find the expected value, i.e., and the average number of people who will lose weight.

(d)   Find the standard deviation \sigma of this deviation.

Solution: Binomial distribution with n=40,p=0.7.

(a) P(X <=30)= 1- P(X < 29) =binomialcdf(40,0.7,29)=.30874

(b)P(16 <=X <= 30) = P(X <=30) –P(X<=15)

=binnomialcdf(40,0.7,30) - binnomialcdf(40,0.7,15) = .80407-0.00002=.80405

(c ) E(X) = np = 40x0.7=28

(e)   \sigma = sqare root of (npq) = square root of 40x0.7x0.3=8.1

[2]The probability to win a certain game is 0.6. What is the probability he wins his first game in the third try? I.e. he loses the first two games and wins the third.

Soltution: Lose the first two games and win the third. Hence there is only one way to achieve this outcome and the probability is (0.4)^2 (0.6)=0.096

[3]A certain area of Caribbean is, on the average, hit by 10 hurricanes a year. Find the probability that for a given year that area will be hit by

(a)      Fewer than 8 hurricanes

(b)   Anywhere between 10 to 12 hurricanes.

Solution: Posisson with \mu=10

(a)    P(X < 8) = P(X <=7) =Poissoncdf(10,7)=0.22022

(b)   P(10 <=X <=12) = P(X <=12) – P(X <=9) = Poissoncdf12)-Poisoncdf(10,9)=.79156-.45792=.33364

[4]A customer service center receives, on the average 6 phone calls an hour. Find the probability that for a given hour that there are

(a)      4 calls

(b)   Between 6 to 8 calls

Solution: Poisson distribution with \mu=6.

(a)    P(X=4) = poissonpdf(6,4)=.18917

(b)   P(6 <=X <=8) =P(X <=8) – P(X<=5)=Poissoncdf(6,8) -Poissoncdf(6,5) =.26574

[5]Suppose that the location of a car accident in LIE West forms a uniform a uniform distribution between 2 km and 6 km from Exit 42. Find the probability the accident happens between 4 km and 6 km from Exit 42. Find the probability density function for this uniform distribution.  Find the probability density function.

Solution: the density function is f(x) = 1/(6-2) = ¼ for 2 < x <6

0,               elsewhere.

P(4 <=X <=6) = (6-4)/(6-2) = ½

[6]The heights of 1000 students are normally distributed with a mean of 175 cm and a standard deviation if 6.9 cm. How many of these students would you expect to have heights

(a)      less than 160 cm?

(b)     between 171 and 182 cm?

(c)      greater than or equal to 188 cm?

(d)     Find the minimum height of a student to be in the upper 70%.

(e)    Find the height k such that P(X < k) = 0.3

Solution: (a) P(X < 160) =normalcdf(-1E99,160,175,6.9) =?

(b) P(171 <= X <= 182) =normalcdf(171,182,175,6.9) =?

(c) P(X >= 188) = normalcdf(188,1E99,175,6.9)

(c)    Find k such that P(X >= k) =0.7.

in otherwords, P(X < k) =0.3 .

Use inversenormal. InvNormal(0.3,175,6.9) =?

(d) Same as (c)

[7]A research scientist reports that mice will live an average of 40 months when their diets are enriched with vitamins. Assume that the life times of such mice are normally distributed with a standard deviation of 6.3 months; find the probability that a given mouse will live

(a)      more than 32 months

(b)     less than 28 months

(c)      between 37 and 49 months?

(d)   How many months do you expect when of 30% of those mice are still living.

Solution: Normal with \mu=40, \sigma=6.3

(a)    P(X >= 32) =normalcdf(32,1E99,40,6.3) =?

(b)   P(X <= 28) = normalcdf(-1E99,28,40,6.3)=?

(c)    P(37 <= X <= 49) =normalcdf(37,49,40,6.3)=?

(d)   find k such that P(X <= k)=0.3.

invnormal(0.3,40,6.3)=?

[8]Find the value z if the area under Consider a normal distribution with =40, =6.3 and its a normal curve.

(a)      Find the area under the curve between z=37 and z=49

(b)     Find the value z whose area under the normal curve to the left of z is 0.3

Solution: Same as [7] (c) and (d)

[9]Draw a standard normal curve. Indicate the maximum point and inflection point.

Solution: It is the bellshpape figure as in the book with inflection point at –1 and +1 and the heigt at x=0 is 1/(sqrare root of 2 \pi)

[10]  (1) Describe the situation where one can use normal distribution as an approximation of binomial distribution.

Describe the situation where one can use Poisson distribution as an approximation of binomial distribution.

(2)Judy found that normal distribution and Poisson distribution are magic and she likes to use them as much as possible. She has two binomial situations. Decide which distribution she should use. Indicate the necessary parameter, erg mean, standard variation.

Situation A: 2000 people are tested. Probability of infection is 0.05.

Situation B: 2000 people are tested. Probability of gaining weight is 0.4

Solution: Generally, when n is large and the probability of success p is close to .5

Generally, n is large and p is small.

Situation A: Can use Poisson with \mu = np = 2000x0.05=100

Situation B: May use normal with

\mu=(2000)(.4)=800,

\sigma = square root of npq =square root of 2000(.4)(.6)