Whenever p 0.5 the binomial distribution will

Please read each question carefully and answer it completely. “Extra work” not explicitly needed can be copied into the end of document in the Appendix. If an answer is incorrect, including work in the Appendix might allow for minimal “partial credit.” (Note that although you are not always required to “show work” on the homework assignments, you may be required to do so on exams.)

#1. Consider the following four continuous probability distributions:

1. Uniform (Continuous) on the interval [0,2]

2. Standard Logistic

3. Standard Normal

4. Exponential with lambda = 2

Please complete the following table with the probability of each event. Please report any decimal answers to three digits.

{0

{X<0.5}

{X=0}

{0.25

1.

0.5

0.25

0

0.25

2.

0.5

0.2441

0

0.304

3.

0.3413

0.6915

0

0.1747

4.

0.8647

0.6321

0

0.3834

Note that for 1. and 2. you need to use techniques other than the distribution commands in EXCEL to find these values. For 1. I would suggest that you draw the distribution and compute the area “geometrically.” You can also use the general form of the cdf in the “Introduction” ppt. For 2. you should use the function for the cdf which you can find either in the .pdf notes or the “Introduction” ppt. For 3. and 4. you should be able to use distribution commands in EXCEL.

#2. Suppose the probability density function for a random variable X equals the following:

f(x) = 3x2 for {0 < x < 1}, f(x) = 0 otherwise.

(a) Show that this is a valid pdf. (Hint: please refer to the two necessary conditions of a pdf from the video introduction and .ppt.)

Condition 1) f(x) is positive everywhere in the support S, that is, f(x) > 0, for all x in S

This condition is true in our case of f(x)=3x^2. f(x) is always positive and greater than zero for all cases.

Condition 2) The area under the curve (i-e) total probability is equal to 1

To find the area under the curve integrate f(x) between limits

(b) Using integration, Find E(X) and the cdf, F(X).

(c) Use F(X) to find the P(0.25 < X < 0.50).

#3. Textbook, p. 293, #45. Question begins: “Is lack of sleep causing traffic fatalities? A study conducted …” Please complete all parts (a-c). On your own, please solve these questions *both* using the standard normal distribution (Table 1) and in EXCEL so that you feel comfortable with both techniques. To formally answer these questions, please fill in the following table using cdf notation for the normal distribution F(x) and standard normal distribution Φ(x), your commands in EXCEL and your final answer. I have started question (a) to demonstrate what I mean, please finish (a) and complete (b) and (c).

1.

Thus the probability that number of fatal crashes will be fewer than 1000 is 0.033401.

1.

Thus the probability that number of crashes will be between 1000 and 2000 is 0.899792.
1.

Thus there should be atleast 2043 fatal crashes for the year to be in top 5%.

cdf notation

EXCEL commands

Answer (for probabilities, three decimal places)

(a)

F(1000) or

Φ(-1.833)

=NORM.DIST(1000,1550,300,TRUE)

=NORM.DIST(-1.833,0,1,TRUE)

or for standard normal:

=NORM.S.DIST(-1.833,TRUE)

0.033

(b)

F(2000)-F(1000)

=NORM.DIST(2000,1550,300,TRUE) -NORM.DIST(1000,1550,300,TRUE)

0.900

(c)

NA

=NORM.INV(0.95,1550,300)

2043

In addition, please complete this question:

d. Using the “Empirical Rule” (pp. 126-127 and p. 274), please fill in the following:

The probability that X is between ___1252___ and __1848______ is approximately 0.68 (or 68%)

The probability that X is between __962____ and ____2138____ is approximately 0.95 (or 95%)

The probability that X is between ___777____ and ____2323_____ is approximately 0.99 (or “almost all”).

#4. Note that there can be an interesting relationship between the Poisson distribution (discrete) and exponential (continuous) distributions, for example: If the number of occurrences per time follows a Poisson distribution, the time between occurrences follows an exponential distribution. [See pp. 288-289].

Say that a bank is conducted a study on the efficiency of its bank tellers and finds that the number of customers served by one teller during an hour period follows a Poisson distribution with a mean of 12. (Assume that there are also customers in line so that the tellers will be continually serving customers.)

Where appropriate, include cdf notation to show the probability you are solving. You can use F(x) for either the Poisson or the exponential distributions.

Hint: this question can be tricky because of the use of µ and λ for the exponential. BE CAREFUL with this distinction. Also, make sure for (d) and (e) you convert minutes to hours for consistency. For example, 15 minutes is 0.25 hours.

(a) What is the probability that one teller will be able to serve more than 10 customers in an hour?

(b) What is the probability that one teller will be able to serve between 3 and 8, (), customers in an hour?

(c) If you observe that a teller, on average, can serve 12 customers an hour, what is the average time it takes for a customer to be served? [Express this average in terms of hours.]

(d) Using your average from (c), use an exponential distribution to find the probability that it will take more than 15 minutes to serve a customer.

(e) Find the probability that it will take between 5 and 10 minutes for a customer to be served.

#5. Textbook, p. 286, #29. Question begins: “An Internal Revenue Oversight Board …” Please complete parts a-c, make sure to apply the continuity correction where necessary. Where appropriate, use pdf/cdf notation to show the probability you are solving.

a) P(X >=6) = P(X =6) + P(X=7) + P(X =8)

= 0.2758+0.3590+0.2044

= 0.8392

b) Using normal approximation, mean = 80*0.82 = 65.6 and standard deviation = Sqrt(80*0.82*0.18) =3.4363

P(X >= 60) = P(X > 59.5) = P(Z > (59.5-65.6)/3.4363))

= P(Z > -1.775)

= 1 - P(Z < -1.775)

= 1 - 0.0379

= 0.9621

c) The binomial distribution is symmetric (like the normal distribution) whenever p = .5. When p ≠ .5 the binomial distribution will not be symmetric. However, the closer p is to .5 and the larger the number of sample observations n, the more symmetric the distribution becomes. On the other hand, the larger the number of observations in the sample, the more tedious it is to compute the exact probabilities of success by use of Equations. Fortunately, though, whenever the sample size is large, the normal distribution can be used to approximate the exact probabilities of success that otherwise would have to be obtained through laborious computations.

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