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1 . The ages X (in years) of n = 200 randomly selected individuals are summarized in the table below. Answer each of the following; show all work. Midpoints x i Age Interval Frequency Relative Frequency Density [0, 2) width = /200 = /2 =.05 3 [2, 4) width = /200 = /2 =.0 8 [4, 2) width = /200 = / [2, 22) width = /200 = /0 = [22, 34) width = /200 = /2 =.025 (a) In the space below, sketch the corresponding density histogram for this random sample. Clearly label all relevant features. (5 pts) PROBLEM CONTINUES ON NEXT PAGE

2 (b) What proportion of this sample can be estimated to be between 9 and 30 years old? Show all work. (8 pts) The 3-year width of the interval [9, 2) is equal to 3/8 the width of the 8-year interval [4, 2); hence the rectangular strip of area above it must be equal to 3/8 of the.30 area of the corresponding class rectangle, or.25. The area of the class rectangle over [2, 22) is equal to.25. The 8-year width of the interval [22, 30) is equal to 8/2 = 2/3 of the width of the 2- year interval [22, 34); hence the rectangular block of area above it must be equal to 2/3 of the.5 area of the corresponding class rectangle, or.0. Therefore, the total proportion is equal to =.4625 of the sample. (c) Calculate the following summary statistics for these grouped data. Show all work. (4 pts ea) Quartile ages Q, Q 2, Q Q 2 2 First, by definition, the median is the value Q 2 that divides the total area of the sample into 0.5 below it, and 0.5 above it. It must therefore lie in the third class interval [4, 2), and divide its class rectangle of area.30 into left (+ right) subareas of.20 (+.0), respectively. Thus the width of the interval [4, Q 2 ) must comprise.20/.30 = 2/3 of the 8-year width of the class interval [4, 2), or 6/3 = years, so Q = years. 2 By definition, the first quartile is the value Q that divides the total area of the sample into 0.25 below it. It must therefore lie in the second class interval [2, 4), and divide its class rectangle of area.20 into left (+ right) subareas of.5 (+.05), respectively. Thus the width of the interval [2, Q ) must comprise.5/.20 = 3/4 of the 2-year width of the class interval [2, 4), or.5 years, so Q = 3.5 years. Similarly, by definition, the third quartile is the value Q 3 that divides the area of the sample into 0.25 above it. It must therefore lie in the fourth class interval [2, 22), and divide its class rectangle of area.25 into (left +) right subareas of (.5 +).0, respectively. Thus the width of the interval [Q 3, 22) must comprise.0/.25 = 2/5 of the 0-year width of the class interval [2, 22), or 4 years, so Q = 8 years. 3 Mean age x using midpoints of class intervals x = [( )(20) + ( 3 )(40) + ( 8 )(60)+ ( 7 )(50)+ ( 28 )(30)] =.55 years 200 Set up BUT DO NOT EVALUATE an expression for the sample variance s 2 = 99 [(.55)2 (20) + (3.55) 2 (40) + (8.55) 2 (60) + (7.55) 2 (50) + (28.55) 2 (30)] 2 s.

3 2. Three birds A, B, and C are perched on a branch. The following information is known: (i) There is a 0.6 probability that A will fly away if B flies away. PAB ( ) = 0.6 (ii) There is a 0.5 probability that B will fly away if C flies away. PB ( C ) = 0.5 (iii)there is a 0.3 probability that C will fly away if A flies away. PC ( A ) = 0.3 (iv) If any two birds fly away, there is a 0.8 probability that the remaining bird will fly away. P( A B C) = P( B A C) = P( C A B) = 0.8 (v) There is a 0.2 probability that all three birds will fly away. PA ( B C ) = 0.2 (a) Calculate the probability of each of the following events. Show all work. (3 pts ea) A and B fly away A and C fly away B and C fly away P( C A B) = P( C A B) P( A B), i.e., 0.2 = 0.8 PA ( B) via (iv) and (v). Hence 0.2 PA ( B ) =, i.e., PA ( B ) = 0.5, and likewise, PA ( C ) = 0.5 and PB ( C ) = (b) Calculate the probability of each of the following events. Show all work. A flies away B flies away C flies away PA ( C) = PC ( APA ) ( ), i.e., 0.5 = 0.3 PA ( ) and (iii). Hence PA ( B) = PA ( BPB ) ( ), i.e., 0.5 = 0.6 PB ( ) and (i). Hence PB ( C) = PB ( CPC ) ( ), i.e., 0.5 = 0.5 PC ( ) and (ii). Hence 0.5 PA= ( ) = PB ( ) = = PC ( ) = = (3 pts ea) (c) Construct a Venn diagram (including all probabilities) of the events A, B, and C. Show all work. (8 pts) A. 50.8* = via (v) * * C =.2 PA ( B C) = 0.28 B. 25.8* =.07 (d) Are any of the following pairs of events statistically independent? Formally justify your answers. A and B A and C B and C (2 pts ea) PAB= ( ) 0.6 via (i), PC ( A ) = 0.3 via (iii), PBC ( ) = 0.5 via (ii), PA= ( ) 0.5. PC ( ) = 0.3. PB ( ) = Unequal, therefore NO. Equal, therefore YES! Unequal, therefore NO.

4 (e) Calculate the probability of each of the following events. Show all work. At least one bird flies away P(At least one) = P(None) = 0.28 = (2 pts ea) Exactly one bird flies away P(Exactly one) = =.5 Exactly two birds fly away P(Exactly two) =.09 Exactly two birds fly away, given that at least one bird flies away P(Exactly two At least one) = redundant, since " Exactly two" is a subset of " At least one" P( Exactly two At least one ).09 = = 0.25 P( At least one).72

5 3. Bob watches the weather report every morning before leaving for work, to see if it is expected to rain later in the day. He also makes his own assessment by observing the morning sky, to see if it is overcast. After many such days, Bob records the following information: (i) The probability that the morning sky was overcast, given that it had rained on a randomly selected day, was 70%. P (Overcast Rain) = 0.7 (ii) The probability that the morning sky was overcast, given that it had not rained on a randomly selected day, was 0%. P (Overcast No Rain) = 0. According to the weather report on a particular morning, the prior probability of rain is 50%. P (Rain) = 0.5 (a) Suppose Bob looks out his window, and sees that the sky is overcast. Calculate the posterior probability that it rains that day. P(Overcast Rain) P(Rain) Via Bayes Law P(Rain Overcast) = P(Overcast Rain) P(Rain) + P(Overcast No Rain) P(No Rain) (0.7)(0.5) 0.35 = = = (0.7)(0.5) + (0.)( 0.5) 0.40 Compare the reported prior probability with this corresponding posterior probability. Specifically discuss how the probability of rain that day is affected by observing that the morning sky is overcast. In the event that the morning sky is overcast, the 0.5 prior probability of rain increases to a posterior probability of (b) Suppose Bob looks out his window, and sees that the sky is not overcast. Calculate the posterior probability that it rains that day. P(Not Overcast Rain) P(Rain) P(Rain Not Overcast) = P(Not Overcast Rain) P(Rain) + P(Not Overcast No Rain) P(No Rain) ( 0.7)(0.5) 0.5 = = = 0.25 ( 0.7)(0.5) + ( 0.)( 0.5) 0.60 Compare the reported prior probability with this corresponding posterior probability. Specifically discuss how the probability of rain that day is affected by observing that the morning sky is not overcast. In the event that the morning sky is not overcast, the 0.5 prior probability of rain decreases to a posterior probability of (c) Imagine that Bob had made a recording error, and that the two probabilities in (i) and (ii) above were actually equal. What assertion could be made about how the probability of rain is affected by observing whether the morning sky is overcast or not, and why? Be as specific as possible. There is no effect. If P(Overcast Rain) = P(Overcast No Rain), it then follows that the events Overcast and Rain are statistically independent. [See Prob 3.5/22(a).] Alternatively, a formal Bayes calculation would result in the conclusion that any (prior) probability P(Rain) is equal to the (posterior) probability P(Rain Overcast), which is the very definition of statistical independence.