File - Jenne Meyer PhD

Week 4 Dr. Jenne Meyer

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In order of last name            

Miracle Brandi Sandy LeAndrew Bren Christine Brandon Maria Rose Monique Michelle Jodi

Discrete Variable – each value of X has its own probability P(X).  Continuous Variable – events are intervals and probabilities are areas underneath smooth curves. A single point has no probability.  Total area under curve = 1 

Defined by two parameters, m and s Almost all area under the normal curve is included in the range m – 3s < X < m + 3s

All normal distributions have the same shape but differ in the axis scales. m = 42.70mm s = 0.01mm

Diameters of golf balls McGraw-Hill/Irwin

m = 70 s = 10

CPA Exam Scores © 2007 The McGraw-Hill Companies, Inc. All rights reserved.

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Since for every value of m and s, there is a different normal distribution, we transform a normal random variable to a standard normal distribution with m = 0 and s = 1 using the formula: x–m z= s

Appendix allows you to find the area under the curve from 0 to z.

Now find P(Z < 1.96): .5000

.5000 - .4750 = .0250

Now find P(-1.96 < Z < 1.96). Due to symmetry, P(-1.96 < Z) is the same as P(Z < 1.96).

.9500

So, P(-1.96 < Z < 1.96) = .4750 + .4750 = .9500 or 95% of the area under the curve.

Some important Normal areas:

Suppose John took an economics exam and scored 86 points. The class mean was 75 with a standard deviation of 7. What percentile is John in (i.e., find P(X < 86)?

86 – 75 = 11/7 = 1.57 x – m zJohn = = 7 s So John’s score is 1.57 standard deviations about the mean.

Suppose John took an economics exam and scored 86 points. The class mean was 75 with a standard deviation of 7. What percentile is John in (i.e., find P(X < 86)?

Suppose John took an economics exam and scored 86 points. The class mean was 75 with a standard deviation of 7. What percentile is John in (i.e., find P(X < 86)?

normal distribution p(lower) p(upper) z x mean std.dev .9420 .0580 1.57 86 75 7

Suppose John took an economics exam and scored 86 points. The class mean was 75 with a standard deviation of 7. What percentile is John in (i.e., find P(X < 86)? John is approximately in the 94th percentile

For example, let m = 2.040 cm and s = .001 cm, what is the probability that a given steel bearing will have a diameter between 2.039 and 2.042cm?  In other words, P(2.039 < X < 2.042)  Excel only gives left tail areas, so break the formula into two, find P(X < 2.039) and P(X < 2.042), then subtract them to find the desired probability: 

P(X < 2.042) = .9773

P(X < 2.039) = .1587

P(2.039 < X < 2.042) = .9773 - .1587 = .8186 or 81.9%

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suppose we wanted the probability of selecting a foreman who earned less than $1,100. In probability notation we write this statement as P(weekly income < $1,100).

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suppose we wanted the probability of selecting a foreman who earned less than $1,100. In probability notation we write this statement as P(weekly income < $1,100).

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=.8413

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suppose we wanted the probability of selecting a foreman who earned less than $1,100. In probability notation we write this statement as P(weekly income < $1,100).

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=.8413

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The mean of a normal probability distribution is 500; the standard deviation is 10. a. About 68 percent of the observations lie between what two values? b. About 95 percent of the observations lie between what two values? c. Practically all of the observations lie between what two values?

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The mean of a normal probability distribution is 500; the standard deviation is 10. a. About 68 percent of the observations lie between what two values? b. About 95 percent of the observations lie between what two values? c. Practically all of the observations lie between what two values? a. 490 and 510, found by 500 +/- 1(10). b. 480 and 520, found by 500 +/- 2(10). c. 470 and 530, found by 500 +/- 3(10).

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A normal distribution has a mean of 50 and a standard deviation of 4. a. Compute the probability of a value between 44.0 and 55.0. b. Compute the probability of a value greater than 55.0. c. Compute the probability of a value between 52.0 and 55.0.

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a. 0.8276: First find z -1.5, found by (44 - 50)/4 and z = 1.25 = (55 - 50)/4. The area between -1.5 and 0 is 0.4332 and the area between 0 and 1.25 is 0.3944, both from Appendix D. Then adding the two areas we find that 0.4332 + 0.3944 = 0.8276.

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b. 0.1056, found by 0.5000 - 0.3994, where z = 1.25.

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c. 0.2029: Recall that the area for z = 1.25 is 0.3944, and the area for z = 0.5, found by (52 - 50)/4, is 0.1915. Then subtract 0.3944 - 0.1915 and find 0.2029.

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Next weeks assignments.  Textbook Assignment: Complete Chapter 6 problems 17,

18, 19 (a, b, c), 20  Textbook Assignment: Complete Chapter 7 problems 11, 13  Moved Final Checkpoint to week 6