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100

  1. P(X>80).

  2. 𝑃(𝑋≤80)

  3. Find the CDF, and explain if it is a valid pdf. 

  1. P(81)= 0.10

  2. 𝑃(𝑋≤80)= 1-P(81)= 1-0.10 

  3. 0.15+0.15+0.20+0.40+0.10 = 1 

Yes, valid because cdf=1










100

Let the distribution of x be as follows: 

  1. Find cdf.

  2. Find P(X<0.5) 

  3. Find E(X) 

  1. [∫4x^3 ] from 0 to 1= 1  

  2. [∫4x^3 ] from 0 to 0.5= (0.5)^4

  3. [∫4x^3*x ] from 0 to 1 OR 

 [∫4x^4 ] from 0 to 1 = 4/5

100

Assume that X is normally distributed with mean 6 and standard deviation equal to 3.0.The random variable Y is also distributed normally with mean -2 and standard deviation equal to 2.5. Assume that X and Y have covariance equal to 1.5.

Derive the distribution of the random variable (X - 2Y).

Z = X-2Y

E(Z) = E(X) - 2*E(Y) = 6 - [2*(-2)] = 10

V(Z) = a^2*V(X) +b^2*V(Y) - 2*a*b*COV(XY) 

= 1*V(X) + 4*V(Y) - 4*COV(X,Y)

= 9 + 6.25*4 - 4*1.5 = 28

100

If you know the population mean (μ) and population variance (σ2), what is the sample mean and sample variances? What are the differences between sample mean and sample variances from their respective population distribution?

What happens to the sample variance as n increases?

For normal distribution:

Population distribution 

  • E(X) = μ

  • Var(x) = σ2

Sample distribution

  • E(X) = μ

  • Var(x) = σ2/n

The difference lies in the sample variance which is equal to the population variance divided by n. There is no difference in the population mean and sample mean

As n increases, the sample variance decreases.


100

What kind of distribution is this? 

  1. A market researcher is interested in the proportion of customers who enter a store and make a purchase. Each customer's decision to buy or not buy is independent of another. What type of distribution should the researcher use to model the probability of exactly three customers making a purchase out of ten?

  2. A factory operates machines that are programmed to cut rods into pieces that are between 10 cm and 15 cm long. Assuming the cutting process is completely random and that any length within this range is equally likely, what is the probability that a randomly selected rod is between 12 cm and 14 cm?

  3. Suppose a dice manufacturer tests the quality of their dice by rolling each one 60 times and recording the outcome. Each die is rolled under identical conditions and the results are assumed to be independent of each other. The quality control process requires finding the likelihood that a given die will land on six 10 times out of 60 r

  1. Binomial 

  2. Uniform 

  3. IID 

200

  1. Mean 

  2. Standard deviation 

  1. 77(0.15) +78(0.15)+79(0.20)+80(0.40)+81(0.10)= 79.15

  2. (77-79.15)^2(0.15)+.......+(81-79.15)^2 (0.10)= 1.5275

SD = sqrt(1.5275)

200

Let the distribution of x be as follows: 

  1. Find Var(x).

  2. Find the median. 

  1. Use ∫ (⅘ -x)^2 f(x) 

Or E(X^2) - (E(X))^2= 2/75 


  1. [∫4x^3*x ] from 0 to m = 0.5 

Find m. m= (0.5)^¼

200

A set of wrist watch prices are normally distributed with a mean of 76 dollars and a standard deviation of 10 dollars. 

What proportion of wrist watch prices are between \  66 dollars and 86 dollars?

Approx 68%

200

In the population, IQ (“intelligence quotient”) scores are normally distributed with a mean of 100 and a standard deviation of 15. Suppose that you plan to obtain IQ scores for a random sample of 20 individuals, from which you will calculate the sample average and the sample standard deviation.

You are given that for the standard normal variable, CDF(1/3) = 0.6305587 and CDF(1.4907) = 0.93198144

a) For any given individual, what is the probability that IQ is greater than 105?

b) For a sample of 20 individuals, what is the probability that the sample average is greater than 105?

#Part a

> 1 - pnorm(105, mean = 100, sd = 15)

[1] 0.3694413

The statistics version of this is:

Use P(IQ>X) = P((X-μ/σ) > Z)

P(IQ>105) = P((105-100)/15 > Z) = 1 - P((105-100)/15 < Z) = 1 - CDFz(1/3) = 1 - 0.6305587 = 0.3694413

#Part b

> 1-pnorm(105, mean = 100, sd = 15/sqrt(20))

[1] 0.06801856

Use P(IQ>X) = P((X-μ)/(σ/sqrt(n)) > Z)

P(IQ>105) = P((105-100)/(15/sqrt(20)) > Z) = 1 - P((105-100)/(15/sqrt(20)) < Z) = 1 - CDFz(1.4907) = 1 - 0.93198144 = 0.06801856

** NOTE THE DIFFERENCE IN STANDARD DEVIATION

200

Suppose you have a fair three-sided die, where each side is labeled with a different number from 4 to 6. Each side has an equal probability of landing face up:

P(X=4) = P(X=5) = P(X=6) = 1/3.


Consider a random sample of two independent throws, where X1 and X2 are i.i.d. random variables representing the outcome of each throw with the probability distribution given above. You are interested in the average value of these two throws, denoted by µ. What is the probability distribution of µ in this scenario? (E.g. If µ =µi, what is the probability of P(µi) where you write that out for all µi,)


300

Let X denote the number of times a photocopy machine will malfunction: 0, 1, 2, or 3 times, on any given month. Let Y denote the number of times a technician is called on an emergency call.

  1. Find Fxy(1,1) (CDF) 

  2. Find PMFs of Y

  3. Find E(Y) 

  4. What is the conditional distribution of X given Y = 2?

  1. 0.15+0.30+0.05=0.15 = 0.65

    1. X=0: 0.2

  2. X=0: 0.2, X=1: 0.5
    X=2: 0.2, X=3: 0.1
    Y=0: 0.5, Y=1:0.3
    Y=2: 0.2

  3. E(Y) = 0(0.5)+1(0.3)+2(0.2) = 0.7

  4. X=0: 0/0.20
    X=1: 0.05/0.20
    X=2: 0.10/0.20
    X=3: 0.05/0.20


300


  1. Find K 

  2. Find P(X=0.5) 

  3. Find P( X>0.5)

  1. K=⅕  or 0.2 

  2. 0

  3. (∫ ½ x) from 0.5 to 1 + (∫⅕ x^3) from 1 to 2 

300

Assume X is normally distributed with mean 3 and variance 15. Let Y = 2X + 1. We also know that P(Y<=5.5) = 0.423. 

Write out the R code to determine the value of q such that P(5.5 <= Y <= q) = .55. Bonus points if you can tell us what that value of q is!

X ~ N(3,15)

Because Y = 2X+1,

E[Y] = 2*E[X] + 1 = 7

V(Y) = b^2 * V(X) = 2^2  * 15 = 60

Thus, Y~N(7, 60)

P(Y<=5.5) = 0.423 is given

Thus, P(Y<=q) = 0.423 + 0.55 = 0.973

Thus, to find q, we can use qnorm (0.973, 7, sqrt(60)) = 21.93

300

The amount of money spent by tourists when visiting Disney World is normally distributed with a standard deviation of $43. How large of a sample must you take in order to ensure that the probability of 0.9 of getting a sample mean within $10 (either larger or smaller) of the population mean?

We are also given the following that qnorm(.95) = 1.645

X ~ N(μ, (43)^2)

P(|x̄-μ| < 10) = .9

P(-10< x̄-μ < 10) = .9

P[ -10/sqrt((43^2)/n) < x̄-μ < 10/sqrt((43^2)/n) ] = .9

Therefore  (x̄-μ) / sqrt((43^2)/n ~ N(0,1)

We are given qnorm(0.95) = 1.645

Because we are finding the probability of 0.9 of getting a sample mean within $10 (note it is two-sided, so we use qnorm(.95),

10 / sqrt((43^2)/n = qnorm(0.95) = 1.645

Thus, 

sqrt(n) = (1.645)(43) / 10 = 7.0735

Thus, n = 50

300

Suppose that 75 percent of the people in a metropolitan area live in the city and 25 percent live in the suburbs. 

a) If 1200 people attending a concert represent a random sample from the metropolitan area, what is the probability that the number of people from the suburbs attending the concert will be 270 ? Focus on setting the formula.


b) Also, what does this R code provide regarding this problem?

> pbinom(269, 1200, 0.75)

[1] 0.02

This is a binomial distribution with n = 1200 and p = 0.25. Thus X ~ Binomial (1200, 0.25)

a.) To find the probability that 270 people will attend)

P(X=270) = (1200 C 270)*0.25^(270)*0.75(1200-270) = 0*0*0 = 0

b.) This represents the probability that the number of people from the city attending the concert will be less than 270.

400

Let X denote the number of times a photocopy machine will malfunction: 0, 1, 2, or 3 times, on any given month. Let Y denote the number of times a technician is called on an emergency call.

  1. Are A and B independent? 

  2. The photocopy shop makes 50,000 copies a month. Each copy is sold for 50 cents each. Every time the machine malfunctions a month, the shop loses $1,000 in profits. On the other hand, everytime a technician is called, the service fee is $500. Let R denote the profits gained in one month. Write an equation for R. (Profit = revenue- losses)

  3.  Find E(R).

  1. (0.2)(0.5)= 0.01 which is not equal to 0.15. Thus, not independent. 

  2. R = 50,000(0.5) - 1000(X)- 500(Y) 

  3. E(R) = 25,000- 1000 E(X) - 500 E(Y) 

25,000- 1000(1.2)-500(0.7)=  $23,450

400


  1. Find E(X) 

  2. Find Var(X) 

  1. (∫ ½ x^2) from 0  to 1 + (∫⅕ x^4) from 1 to 2 

  2. Use ∫ (µ-x)^2 f(x) Or E(X^2) - (E(X))^2 

400

We are interested in the earnings of working husband and wife pairs. Assume that the average annual earnings of husbands is normally distributed with mean $42,000 and standard deviation 15,000. The average annual earnings of the wives is also normally distributed with mean $25,000 and standard deviation 8,000. 

Assume that the annual earnings of husbands and wives are independent. Given that the earnings of the wife are greater than $20,000, what is the probability that the husband's earnings are less than $40,000? 

You can simplify $42,000 to 42 when calculating. You do not need to solve this problem entirely, but try to provide the correct R code

H ~ Normal (42, (15)^2) and W ~ Normal (25, (8)^2)

** H and W are independent **


The questions asks to find P(H<40 | W>20) = P(H<40) because H and W are independent.

P(H<40) = pnorm (40, 42, 15) = 0.447

pnorm  is written by (cdf value, mean, standard dev)

400

The filling machine at a bottling plant is operating correctly when the variance of the fill amount is equal to 0.3 ounces. Assume that the fill amounts follow a normal distribution.  Assuming the bottling plant is operating correctly, what is the probability that for a sample of 30 bottles the sample variance is less than 0.5?

Set up the R code for this problem. In the midterm, you would be given a table that would help you find this answer without code

X ~ random variable represent full amounts

X ~ Normal (μ, .3)

Random sample: X, X, … X  i.i.d. Normal (μ, .3)


SX2 = (1/(n-1))*Σ(Xi-x̄)2

SX2  = 1/(30-1) * Σ(Xi-x̄)2

SX2  = 1/29 * Σ(Xi-x̄)2


P(SX2> 0.5) = P(1/29 * Σ(Xi-x̄)2 > 0.5)

  • = P( Σ(Xi-x̄)2 > 0.5*29)

  • = P ( Σ(Xi-x̄)2/(.3) > 0.5*29/0.3))

  • = P(Z>48.33) where Z ~ X292

pchisq (48.33, df=29) = 0.986

400

Assume that the random variable X is uniformly distributed over the interval [4,8]. 

(a) Derive the cumulative distribution function for X. 

(b) Compute E(X^3 - 2X)

X ~ U(4,8) (recall U(a,b))

(a) Fx(x) = integral from x=4 to x=x0 of 1/4 *dx

= ¼ | from x=x0 to x=4

= ¼*x0 - 1  for 4<x0<8


(b) E[X^3 - 2*X] = integral from x=4 to x=8 of ¼(x^3 - 2x) *dx

= [(x^4)/(16) - (x^2)/(4) | from x=8 to x=4 ]

= 228