Key to Test #3, 12/12/2005

STAtistics / Mr. Hansen
12/12/2005 [modified 12/10/2010]

Name: _________________________

Key to Test #3

Probability, symbolic logic, random variables, LOLN
Note: Simulations were on the list of topics, but most of you noticed that there was no room to squeeze a simulation question onto this test. (However, a few of the make-up tests had a simulation question.) The topic of simulations is fair game on tests as well as for the midterm exam.

Part I: Translations into English.

1.	The expected value of a sample mean equals the population mean.
2.	The expected value of random variable X, also called the mean of X, equals the probability-weighted sum of numeric outcomes. Also acceptable: the sum of products formed by multiplying each numeric outcome by its probability of occurring.
3.	The variance of r.v. X equals the probability-weighted sum of squared deviations from the mean of X.
4.	The conditional probability of event A, given that event B is true, equals the probability that A and B both occur divided by the probability of B.

Part II: Normal Distributions.

5.	continuous

6.(a)	.524 [Look for value closest to .7000 in the body of the z table, or use calculator keystrokes invNorm(.7) ENTER. The invNorm function is the third option on the 2nd DISTR menu.]

(b)	13.36%
	[Your work consists of sketching a normal curve and shading the left and right tails. By normalcdf(1.5,9999), which you cannot write, you get an area of 0.0668 in each tail, which you should mark with two arrows, one pointing to each tail.]

(c)	z* = 2.58 = solve for x to get 893.8 Similarly, use the other z* value, namely, −2.58, to get 326.2. Answer: The most extreme 1% of test-takers are those whose scores are outside the interval of 326 to 894. [Use table of critical values, inside back cover of textbook, to get z*. That means that we have to look 2.58 standard deviations to the left and 2.58 standard deviations to the right to find the cutoff test score values. A sketch of the z curve is helpful but not required in part (c).]

Part III: Short Answer.

7.	B(12, 0.85); i.e., binomial with n = 12, p = 0.85
8.	geometric, n = undefined or DNE, p = 0.85
9.	P(X < 9) = 0.092 = the probability that Carl sinks fewer than 9 free throws in 12 tries [calculator keystrokes: binomcdf(12,.85,8) ENTER]
10.	= 1/p = 1.176 [You are responsible for knowing this formula. It is common sense: 6 tries are needed, on average, to roll a 3 with a fair die since probability is 1/6; 52 tries are needed, on average, to draw the ace of spades from a well-shuffled 52-card deck since probability is 1/52 if deck is reloaded and reshuffled after each attempt.]
	[This formula might be provided, since it is not on the AP sheet.] [If the formula is not provided, you can still answer the question by entering values 1,2,3,4,5,6,7 into list L₁, probabilities .85, .1275, .01913, .00287, .0004303, .00006545, and .000009682 into L₂. Then use STAT CALC 1 L₁,L₂ ENTER to see the mean and s.d. If you can’t figure out how to calculate the probability entries of .85, .1275, etc., then please reread your textbook or call Mr. Hansen in the middle of the night at 703-599-6624, because this is a genuine emergency.]
11.	= np = 12(.85) = 10.2
12.	= 1.237
13.	P(Y > 2) = 0.0225 = the probability that Carl’s first success occurs on or after the third shot [Keystrokes: 1 − geometcdf(.85,2) ENTER.]
14.	P(63 < Z < 68) = 0.673 = the probability that a randomly selected adult American woman is between 63 and 68 inches tall [work consists of a normal curve with shaded area]
15.	No, Z is a normally distributed r.v.
16.	P(at least one crash) = 1 – P(no crashes) = 1 – (0.9999995⁴⁰⁰⁰)³⁶⁵ = 0.518