5: Several Useful Discrete Distributions

Transcription

1 : Several Useful Discrete Distributions. Follow the instructions in the My Personal Trainer section. The answers are shown in the tables below. The Problem k P( k) List the Values of Write the probability Rewrite the probability Find the probability Three or less 0,,, P( ).0 Three or more,,, 6, 7, P( ) - P( ).0 =.99 More than three,, 6, 7, P( > ) - P( ).0 =.9 Fewer than three 0,, P( < ) P( ).0 Between and (inclusive),, P( ) P( ) - P( )..0 =.7 Eactly three P( = ) P( ) - P( ).0.0 =.07. Follow the instructions in the My Personal Trainer section. The answers are shown in the tables below. k P( k) The Problem List the Values of Write the probability Rewrite the probability Find the probability Eactly two P( = ) P( ) - P( ).6.96 =.67 More than two,,,9 P( > ) - P( ).6 =.7 Two or more,,,,9 P( ) - P( ).96 =.0 Less than two 0, P( < ) P( ).96 Between and (inclusive),, P( ) P( ) - P( ) =.70 Two or less 0,, P( ).6. The random variable is not a binomial random variable since the balls are selected without replacement. For this reason, the probability p of choosing a red ball changes from trial to trial.. If the sampling in Eercise. is conducted with replacement, then is a binomial random variable with independent trials, and p = P red ball =, which remains constant from trial to trial. n = [ ] (7) C..7 = =.96 () 6. a ( ) ( ) ( )( ) b 0 C ( ) ( ) = ( ) = c ( 9)( ) ( ) ( ) ( )( ) ( ) C.. =. = d ( ) ( ) ( )( ) C.. = 7.. =.670

2 ! 0!! a C 0(.) (.) = ()(.67) =.67 b ( ) ( ) ( )( ) 6 ( ) ( ) ( )( ) C.. =..097 =. c C.. =.0.6 =.96 d P ( ) = =.0 e P ( ) = = a For 7 and p =., P( = ) = C..7 =.097. n = ( ) ( ) b These probabilities can be found individually using the binomial formula, or alternatively using the cumulative binomial tables in Appendi I. P ( ) = p(0) + p() = C (.) (.7) + C (.) (.7) = (.7) + 7(.)(.7) = =.9 or directly from the binomial tables in the row marked a =. c Refer to part b. P ( > ) = P ( ) =.9 =.67. d np ( ) µ = = 7. =. e σ = npq = 7(.)(.7) =.7 =.. With n = 6 and., the values of = C are calculated for = 0,,,,,, and 6. The values of p() are shown in the table below. 0 6 p() The probability histogram is shown below. 6 p = p ( ) ( p) ( q) 6 0. n = 6, p =. 0. p() Notice that when =., ( ) = (.) (.) 6 6. In Eercise., with., ( ) (.) (.) 6 p p C probability that = k when. --- p = ( ) n k ( ) k p = 6 (.) k (.) C k This follows because C n k n n! n Ck = = Cn k k! n k! ( ) n k p = p = C. The --- is the same as the probability that = n k when Therefore, the probabilities p() for a binomial random variable when n = 6 and p =. will be the mirror images of those found in Eercise. as shown in the table. The probability histogram is shown on the net page. 0 6 p()

3 0. n = 6, p =. 0. p() n n. Refer to Eercise.9. If., ( ) (.) (.) (.) n n n n p = p = C = C. Since C = Cn (from Eercise.9), you can see that for any value of k, n ( ) (.) n n n P = k = Ck = Cn k(.) = P( = n k) This indicates that the probability distribution is eactly symmetric around the center point.n.. a For and p =., P( = ) = C..6 =.. n = ( ) ( ) b To calculate P ( ) = p() + p() + L+ p() it is easiest to write P ( 0) C P ( ) = P ( < ) = P ( ). These probabilities can be found individually using the binomial formula, or alternatively using the cumulative binomial tables in Appendi I. 0 = = ( ) ( ) = P ( ) ( ) 9 ( = ) = C..6 =.00 = = ( ) ( ) = ( ) ( ) 7 P ( ) C..6. P ( = ) = C..6 =. The sum of these probabilities gives P ( ) =.and P ( ) =. =.6. c Use the results of parts a and b. P ( > ) = P ( ) =. +. =.67 ( ) d From part c, P ( ) = P ( ) + P ( = ) =. +. =.6. e np ( ) µ = =. = f σ = npq = (.)(.6) =. =.9. Table, Appendi I gives the sum, p(0) + p() + L + p( k), for various values of n and p. The three necessary sums can be found directly in the table, by indeing the proper values of n, p and k. a p(0) + p() + p() + p() =.97 for n= and p =. b p(0) + p() + L+ p(7) =. for n= and p =.6 c p(0) + p() + L+ p() =.7 for n= and p =.. a P[ ] P[ ] b P[ ] P[ ] P[ ] c P[ < ] = P[ ] =. 00 d P[ ] P[ ] = =.099 =.90 = = = =.0 > = =.00 =.99. Entries can be read directly from Table, Appendi I..9. a P[ ] = b P[ ] = c [ ] P 9 =.6 6

4 . a P[ < ] = P[ ] =.7 b P[ 6 ] =.6 c P[ > ] = P[ ] = = d P[ ] = P[ ] = = e P[ ] P[ ] P[ ] < < 7 = 6 =..7 =.66.6 For a binomial random variable and n identical trials and p P[ success] deviation are µ = np and σ = npq respectively. a µ = 00(.) = 00; σ = 00(.)(.7) =.9 b µ = 00(.0) = ; σ = 00(.0)(.99) =.99 c µ = 00(.) = 0; σ = 00(.)(.) =. d µ = 600(.) = 0; σ = 600(.)(.) = 6 =, the mean and standard.7 a µ = 0(.0) = ; σ = 0(.0)(.99) =.99 b µ = 0(.9) = 90; σ = 0(.9)(.) = c µ = 0(.) = 0; σ = 0(.)(.7) =. d µ = 0(.7) = 70; σ = 0(.7)(.) =. e µ = 0(.) = 0; σ = 0(.)(.) =. The table below shows the values of p and σ (which depends on p) for a given value of n. The maimum value of p occurs when p =.. p σ a p(0) = 0 (.) (.9) =.767 p ( ) ( ) C 7 () = C..9 = = ( ) ( ) = p ( ) ( ) p() C () = C..9 = p() = C (.) (.9) =.79 so that P[ ] = p(0) + p() + p() + p() + p() =.96 b Using Table, Appendi I, P[ ] is read directly as.97. c Adding the entries for = 0,,,,, we have [ ] d µ = np = 0(.) = and σ = npq =. =.6 e For k =, µ ± σ = ±. or.6 to. so that f [ ] [ ] P =.966. P.6. = P = =.7 For k =, µ ± σ = ±.6 or.6 to.6 so that [ ] [ ] P.6.6 = P 0 =.969 For k =, µ ± σ = ±.0 or.0 to 6.0 so that [ ] [ ] P = P 0 6 =.9977 The results are consistent with Tchebysheff s Theorem and the Empirical Rule..0 Although there are n = 0 days on which it either rains (S) or does not rain (F), the random variable would not be a binomial random variable because the trials (days) are not independent. If there is rain on one day, it will probably affect the probability that there will be rain on the net day. 7

5 . Although there are trials (telephone calls) which result in either a person who will answer (S) or a person who will not (F), the number of trials, n, is not fied in advance. Instead of recording, the number of successes in n trials, you record, the number of trials until the first success. This is not a binomial eperiment.. There are n = 0 students which represent the eperimental units. a Each student either took (S) or did not take (F) the SAT. Since the population of students is large, the probability p =. that a particular student took the SAT will not vary from student to student and the trials will be independent. This is a binomial random variable. b The measurement taken on each student is score which can take more than two values. This is not a binomial random variable. c Each student either will (S) or will not (F) score above average. As in part a, the trials are independent although the value of p, the proportion of students in the population who score above average, is unknown. This is a binomial eperiment. d The measurement taken on each student is amount of time which can take more than two values. This is not a binomial random variable.. Define to be the number of alarm systems that are triggered. Then p P[ ] = alarm is triggered =.99 and n = 9. Since there is a table available in Appendi I for n = 9 and p =.99, you should use it rather than the binomial formula to calculate the necessary probabilities. P at least one alarm is triggered = P = P = 0 =.000 =.000. a [ ] ( ) ( ) b P[ ] = P( > ) = P( ) = = c P[ eight or fewer] = P( ) =.06 more than seven Define to be the number of people with Rh-positive blood. Then p P[ ] = Rh-positive =. and n =. Since there is no table available for p =., you will have to calculate the probabilities using the binomial formula. ( ) ( ) 0( ) ( ) ( ) 0 P both Rh-negative = P = 0 = C.. =. =.0. Define to be the number of cars that are black. Then p P[ ] Appendi I. P = P =.90 =.09 a ( ) ( ) b P( 6 ) =.99 c P( > ) = P( ) =.90 =.09 d P( = ) = P( ) P( ) = =. e P( ) = P( ) P( ) = =.0 f P( not ) P( ) P( ) more than 0 black = less than black = =.90.6 Define to be the number of readers or older. Then p P[ ] the binomial formula or Table to calculate the necessary probabilities. P = P =.07 =.9 a [ ] [ ] b P[ ] P[ ] P[ ] c P[ < ] = P[ ] =. 00 = 9 = 9 = =. = black =.and n =. Use Table in = success =.6 and n =. You can use either.7 Define a success to be a patient who fails to pay his bill and is eventually forgiven. Assuming that the trials are independent and that p is constant from trial to trial, this problem satisfies the requirements for the binomial eperiment with n = and p =.. You can use either the binomial formula or Table.

6 0 a P[ ] p() = C ( ) ( ) ( ) = =..7 =. =.00 b P[ ] p() = C ( ) ( ) ( )( ) = =..7 =..7 =.6 0 c P[ ] 0( ) ( ) ( ) = 0 = C..7 =.7 =.0. The mean number of bills that would have to be forgiven is given by µ = np = 000(.) = 600 ; the variance of is σ = npq = 000(.)(.7) = 0 andσ = 0 = It is necessary to approimate [ 700 ]. P > From Tchebysheff s Theorem we know that at least ( k ) of the measurements lie within kσ of the mean. The value = 700 is 0 units away from the mean µ = 600. This distance is equivalent to 0 σ =. standard deviations from the mean. For a point k =. standard deviations from the mean, Tchebysheff s Theorem concludes that at least ( ). =.96 of the measurements lie within.σ of the mean (i.e., 600 ± 0 ). Therefore, at most. 96 =.0of the measurements are less than 00 or greater than 700. Since the distribution is fairly mound-shaped and symmetric ( p =. and n large) we can say that at most.0 but more likely.0 of the measurements are greater than Define to be the number of fields infested with whitefly. Then p P[ ] = infected field =.and n = 0. a µ = np = 0(.) = b Since n is large, this binomial distribution should be fairly mound-shaped, even though p =.. Hence you would epect approimately 9% of the measurements to lie within two standard deviation of the mean with σ = npq = 0(.)(.9) =. The limits are calculated as µ ± σ ± 6or from to 6 c From part b, a value of = would be very unlikely, assuming that the characteristics of the binomial eperiment are met and that p =.. If this value were actually observed, it might be possible that the trials (fields) are not independent. This could easily be the case, since an infestation in one field might quickly spread to a neighboring field. This is evidence of contagion..0 Define to be the number of times the mouse chooses the red door. Then, if the mouse actually has no p = P red door =. and n =. Since µ = np = and σ = npq =., you would preference for color, [ ] epect that, if there is no color preference, the mouse should choose the red door µ ± σ ±.6. to.6 or between and times. If the mouse chooses the red door more than or less than times, the unusual results might suggest a color preference.. Define to be the number of travelers listing traffic and other drivers as their pet peeve. Then, n = and p =.. 0 a P( = ) = C (.) (.67) = b From the Minitab Probability Density Function, read P( = ) =.000. c Use the Minitab Cumulative Distribution Function to find P( 7 ) = Define to be the number of Americans who look for services close to the highway. Then, n = and p =.. a µ = np = (.) = and σ = npq = (.)(.6) =.9 b µ ± σ ±.9. to.9. Since can take only integer values from 0 to, this interval consists of the values of in the range 6. 9

7 c Using Table in Appendi I, P( ) P( ) P( ) 6 = = =.97. This value agrees with Tchebysheff s Theorem (at least / of the measurements in this interval) and also with the Empirical Rule (approimately 9% of the measurements in this interval).. Define to be the number of Americans who are tasters. Then, n = 0 and p =.7. Using the binomial tables in Appendi I, a P ( 7) = P ( 6) =.9 =.7 b P ( ) =.76. Define to be the number of households that have at least one dog. Then, n = and p =.. Using the binomial tables in Appendi I, a P ( = ) = P ( ) P ( 7) = =. b P ( ) =.7 c P ( > ) = P ( ) =.99 =.009. Follow the instructions in the My Personal Trainer section. The answers are shown in the table below. Probability Formula Calculated value P( = 0) 0.. e.0 0! P( = ).. e.0! P( = ).. e.6! P( or fewer successes) P( = 0) + P( = ) + P( = )..6 Follow the instructions in the My Personal Trainer section. The answers are shown in the table below. Probability Formula Calculated value P( = 0) 0 e.09 0! P( = ) e.9! P(more than one success) [P( = 0) + P( = )].00.7 Follow the instructions in the My Personal Trainer section. The answers are shown in the tables below. k P( k) The Problem List the Values of Write the probability Rewrite the probability Find the probability Three or less 0,,, P( ) Three or more,, P( ) - P( ). =.77 More than three,, P( > ) - P( ).67 =. Fewer than three 0,, P( < ) P( ). Between and (inclusive),, P( ) P( ) - P( ).96. =.9 Eactly three P( = ) P( ) - P( ).67. =. 0

8 . Follow the instructions in the My Personal Trainer section. The answers are shown in the tables below. k 0 P( k) The Problem List the Write the Rewrite the Find the Values of probability probability probability Eactly two P ( = ) P ( ) P ( ).9.09 =. More than two,, P> ( ) P ( ).9 =.07 Two or more,, P ( ) P ( ).09 =.9 Less than two 0, P ( < ) P ( ).09 Between and (inclusive),, P( ) P ( ) P ( ) =.90 Two or less 0,, P ( ).9 µ µ e e.9 Using p ( ) = =,!! 0 e a P[ = 0 ] = =. 0! e = = =.7067! P > = P = =.999 b P[ ] c [ ] [ ] d P[ ] e = = =.0609!.0 Table in Appendi I gives the cumulative probabilities, P[ k] a P[ ] = P[ ] =.9 =.9 b P[ < 6] = P[ ] =. 9 c P[ = ] = P[ ] P[ ] =..7 =.7 d P[ ] = P[ ] P[ 0 ] =.9.0 =.09. a Using Table, Appendi I, P[ ] =.677 µ = = ( ) = p( ) b With np 0., the approimation is c for various values of µ and k.! e. Then 0 0 e e e P[ ] + + 0!!! = =.677 The approimation is quite accurate.. For a binomial distribution with n = and p =.0, the mean is µ = (.0) =.. Using the Poisson distribution with µ =., we can approimate the binomial probabilities required: (.) (.) 0. 0 e p ( 0 ) = C 0 (.0) (.9) =.6 0!. e p ( ) = C (.0) (.9) =.! From Table, Appendi I in the tet with n =, p =.0, we find that the eact probabilities are p(0) =.77 and p() =.6.77 =.6

9 Note the accuracy of the Poisson approimation for n as small as.. Let be the number of misses during a given month. Then has a Poisson distribution with µ =. a e p(0) = e =.0067 b p() = =.7! P = P =.0 =.60 from Table. c [ ] [ ]. a 0 e e e e p() = =.0 and P[ ] = + + =.6! 0!!! b Recall that for the Poisson distribution, µ = and σ = =.6 Therefore the value = lies ( ).6 =.6 standard deviations above the mean. It is not a very likely event.. Let be the number of injuries per year, with µ =. a P[ ] P[ ] P[ ] b P[ ] P[ ] c P[ ] =..06 = = = =.7 = =.06 =.9.6 Refer to Eercise.. a µ = and σ = µ = =. b The value of should be in the interval µ ± σ ±.. to. at least / of the time (and probably more). Hence, most value of will lie between 0 and..7 The random variable, number of bacteria, has a Poisson distribution with µ =. The probability of interest is P eceeds maimum count = P > [ ] [ ] Using the fact that µ = and σ =. from Eercise.7, most of the observations should fall within µ ± σ or 0 to. Hence, it is unlikely that will eceed. In fact, the eact Poisson probability is P[ > ] =.07.. The random variable is the number of cases per 0,000 of E. Coli in a two-year period. If you assume that has a Poisson distribution with µ =.,.9 a ( ) b P( ) P( ) a P =.9 (from Table ) > = =.9 =.0 c With µ =. and σ = µ =., 9% of the values of should lie in the interval µ ± σ. ±.6.66 to.66 That is, 9% of the occurrences involve between 0 and cases, which would be at most cases. C.0 The formula for p ( ) is a () = =.6 b 6 C 7 C p ( ) = for =,,,,. 6() = =. c ().07 C = 70 = Since there are only failures and we are selecting items, we must select at least one success. P = 0 = 0. Hence, [ ] 0

10 6 P = P = p() = = =.976 b [ ] [ ] c P[ ] () = = = =. C 6. The formula for p ( ) is a b C 6 C p ( ) = for = 0,,, 6 0 p(0) = = =.6 () = = =. 0 p C C 66 0 p() = = =. p() = = =.0 C C The probability histogram is shown below p() c Using the formulas given in Section.. M µ = E ( ) = n = =. N M N M N n σ = n =.06 N N N = d Calculate the intervals µ ± σ =. ±.06 =. ±. or.6 to. µ ± σ =. ±.06 =. ±. or.7 to.97 Then, P.6. = p(0) + p() + p() =.99 [ ] [ ] P.7.97 = p(0) + p() + p() =.99 These results agree with Tchebysheff s Theorem.. The formula for p ( ) is p ( ) = for = number of blue candies = 0,,,. C a P( = ) = = =.7 b P( ) c P( ) C 0 C () 6 = = = = The formula for p ( ) is 6 C = 0 = = = C p ( ) = for = number of defectives = 0,,. Then p(0) = = =. p() = = =.6 p() = = = C C C

11 These results agree with the probabilities calculated in Eercise.90.. a The random variable has a hypergeometric distribution with N =, M = and n=. Then b c p ( ) = for = 0,, C Using the formulas given in this section of the tet, M µ = n = = =. N M N M N n 9 σ = n = = =. 6 N N N σ =.6 =.6 The probability distribution and histogram for are shown below. 0 p() / 6/ / p() a The random variable has a hypergeometric distribution with N =, M = and n=. Then p ( ) = for = 0,,, C 0 b P( = ) = = =.76 c P( ) d P( ) C 6 + = + = = C C = 0 = = = The random variable has a hypergeometric distribution with N =, M = and n=. Then p ( ) = for = 0,,,, C 0 C 0 a P( = ) = = = b P( ) P( ) c ( ) C C = = = = + () + () 0 P = p() + p() = = = = 7.7 See Section. in the tet.. The Poisson random variable can be used as an approimation when n is large and p is small so that np < 7. The Poisson random variable can also be used to model the number of events occurring in a specific period of time or space.

12 .9 The hypergeometric distribution is appropriate when sampling from a finite rather than an infinite population of successes and failures. In this case, the probability of p of a success is not constant from trial to trial and the binomial distribution is not appropriate..60 The random variable is defined to be the number of heads observed when a coin is flipped three times. p = P success = P head =, q = p = and n =. The binomial formula yields the Then [ ] [ ] following results. 0 a P[ = 0 ] = p(0) = C 0( ) ( ) = P[ ] p C ( ) ( ) P[ = ] = p() = C ( ) ( ) = P[ ] p C ( ) ( ) b The associated probability histogram is shown below. 0. = = () = = 0 = = () = = 0. p() c µ np ( ) σ npq ( )( ) d 0 = = =. and = = =.66 The desired intervals are µ ± σ =. ±.66 or.6 to.66 µ ± σ = ±.7 or. to. The values of which fall in this first interval are = and =, and the fraction of measurement in this interval will be + =. The second interval encloses all four values of and thus the fraction of measurements within standard deviations of the mean will be, or 0%. These results are consistent with both Tchebysheff s Theorem and the Empirical Rule..6 Refer to Eercise.60 and assume that p =. instead of p =.. 0 a P[ = 0 ] = p(0) = C 0(.) (.9) =.79 P[ ] p C ( ) ( ) P[ = ] = p() = C (.) (.9) =.07 P[ ] p C ( ) ( ) = = () =..9 =. 0 = = () =..9 =.00 b Note that the probability distribution is no longer symmetric; that is, since the probability of observing a head is so small, the probability of observing a small number of heads on three flips is increased (see the figure below) p() c µ np ( ) σ npq ( )( ) = =. =. and = =..9 =.0

13 d The desired intervals are µ ± σ =. ±.0 or.0 to.0 µ ± σ =. ±.0 or.70 to. The only value of which falls in this first interval is = 0, and the fraction of measurements in this interval will be.79. The values of = 0 and = are enclosed by the second interval, so that =.97 of the measurements fall within two standard deviations of the mean, consistent with both Tchebysheff s Theorem and the Empirical Rule..6 It is given that n = 0, p =. and = number of patients surviving ten years. a P[ ] P[ ] b P[ ] P[ ] c P[ ] P[ ] = = 0 =.000= = 9 =. =. = =.979 =.0.6 Define to be the number supporting the commissioner s claim, with p =. and n =. a Using the binomial tables for n P[ ] P[ ] b P[ = ] = P[ ] P[ ] = =.6 =, = =.766 =. c The probability of observing an event as etreme as = (or more etreme) is quite high assuming that p =.. Hence, this is not an unlikely event and we would not doubt the claim..6 a Define to be the integer between 0 and 9 chosen by a person. If the digits are equally likely to be chosen, then p ( ) = for = 0,,, K,9. b P[, or 6 is chosen ] = p() + p() + p (6) = 7 c P [ not, or 6] = =.6 Refer to Eercise.6. Redefine to be the number of people who choose an interior number in the sample of n = 0. Then has a binomial distribution with p =.. a P[ ] P[ ] = 7 =.77 =. b Observing eight or more people choosing an interior number is not an unlikely event, assuming that the integers are all equally likely. Therefore, there is no evidence to indicate that people are more likely to choose the interior numbers than any others..66 Let be the number of families in the sample of n = who have saved $000 or more for college. Then has a binomial distribution with p =.. a p ( ) ( ) ( ) = C.. for = 0,,, K,. b From Table, P( ) =.696. c P( ) P( ) > = =.696 =.0. d From Table in Appendi I, look down the column for p =. to find the largest cumulative probability which is still less than or equal to.. This probability is.09 with Therefore, the largest value of c is c =. P( ) = Let be the number of Americans who rank owning a vacation home as the number one status symbol. Then has a binomial distribution with n= 00 and p =.6. a µ = np = 00 (.6) =0 b σ = npq = 00 (.6)(.) = 9.79 c Since n is large and p =.6, this binomial distribution will be relatively mound-shaped, and approimately 9% of the values of should lie in the interval 6

14 µ ± σ 0 ± to 9.96 d The value 00 lies ( ) = =.0 standard deviations below the mean. This is an unusual result. Perhaps the sample was not randomly selected, or perhaps the 60% figure is no longer accurate, and in fact this percentage has decreased..6 Let be the number of reality TV fans in a sample of n = 0 who say their favorite reality show involves escaping from remote locations. Then has a binomial distribution with p =.. a P( ) P( ) P( ) b P( ) P( ) P( ) P( ) =.0 = 6 = 6 = =.00 = =.979 =.0 c. This would be an unlikely occurrence, since it occurs less than time in It is given that = number of patients with a psychosomatic problem, n =, and p = P patient has psychosomatic problem. A psychiatrist wishes to determine whether or not p =.. [ ] a Assuming that the psychiatrist is correct (that is, p =. ), the epected value of is E ( ) = np= (.) = 0. b σ = npq = = c Given that., (.)(.) p = [ ] P =.006 from Table in Appendi I. d Assuming that the psychiatrist is correct, the probability of observing = or the more unlikely values, = 0,,,, is very unlikely. Hence, one of two conclusions can be drawn. Either we have observed a very unlikely event, or the psychiatrist is incorrect and p is actually less than.. We would probably conclude that the psychiatrist is incorrect. The probability that we have made an incorrect decision is P given p =. =.006 which is quite small. [ ].70 Define to be the number of students favoring the issue, with assumed to be.. Using the binomial tables in Appendi I, P =.07 [ ] n = and p = P[ student favors the issue] Thus, the probability of observing = or the more etreme values, = 0,,,, is quite small under the assumption that p =.. We probably should conclude that p is actually smaller than...7 Define to be the number of students 0 years or older, with 00 a Since has a binomial distribution, np 00(.) 0 b The observed value, =, lies 0 =. 6. standard deviations below the mean. It is unlikely that p =...7 a p P[ ] = rain according to the forecaster =. n = and p P[ ] = student is 0+ years =.. µ = = = and npq ( )( ) b With n= and p =., µ = np = (.) = 7. and σ = npq =. =.99 c The observed value, =, lies σ = = = z = =.09.9 standard deviations above the mean. d The observed event is not unlikely under the assumption that p =.. We have no reason to doubt the forecaster. 7

15 .7 a If there is no preference for either design, then p P[ ] = choose the second design =.. b Using the results of part a and n=, µ = np = (.) =. and σ = npq = 6. =.. c The observed value, = 0, lies 0. z = =. standard deviations above the mean. This is an unlikely event, assuming that p =.. We would probably conclude that there is a preference for the second design and that p >...7 The random variable, the number of neighbors per square meter, has a Poisson distribution with µ =. Use the Poisson formula or Table in Appendi I. P = 0 =.0 a ( ) b P( ) =. c P( ) P( ) = =.69 =.7 d With µ = and σ = µ =, approimately 9% of the values of should lie in the interval µ ± σ ± 0 to In fact, using Table, we can calculate the probability of observing between 0 and neighbors per square meter to be P =.979, which is close to our approimation. ( ).7 a The random variable, the number of plants with red petals, has a binomial distribution with n = p = P red petals =.7. and [ ] b Since the value p =.7 is not given in Table, you must use the binomial formula to calculate ( ) ( ) ( ) ( ) ( ) P 9 = C.7. + C.7. = =.0 ( ) ( ) ( ) ( ) ( ) c P = C.7. + C.7. = d Refer to part c. The probability of observing = or something even more unlikely ( = 0) is very small This is a highly unlikely event if in fact p =.7. Perhaps there has been a nonrandom choice of seeds, or the 7% figure is not correct for this particular genetic cross..76 a The random variable, the number of chickens with blue feathers, has a binomial distribution with 0 and p = P blue feathers = P Bb + P bb =.. n = [ ] ( ) ( ) b µ = np = 0(.) =. c From Table, P( < ) = P( ) =.006. d From Table, P( ) P( ) P( ) = 9 =.6. = Let be the number of lost calls in a series of n = trials. If the coin is fair, then binomial distribution. a P[ ] b If, = = = which is the same as odds of :07. 0 n = P[ ] = = = which is a very unlikely event. 9 p = and has a.7 a Since cases of insulin-dependent diabetes is not likely to be contagious, these cases of the disorder occur independently at a rate of per 0,000 per year. This random variable can be approimated by the Poisson random variable with µ =. b Use Table to find P( ) =.6.

16 c P( ) P( ) P( ) d 7 = 7 =.67. =.7. The probability of observing or more cases per 0,000 in a year is P = P 9 =.96 =.0 ( ) ( ) This is an occurrence which we would not epect to see very often, if in fact µ =..79 a The distribution of is actually hypergeometric, with N = 00, n= 0 and M = number of defectives in the lot. However, since N is so large in comparison to n, the distribution of can be closely approimated by the binomial distribution with 0 p = P defective. n = and [ ] b If p is small, with np < 7, the Poisson approimation can be used. c If there are defectives in the lot, then p = 00 =.00 and µ =.667. The probability that the lot is shipped is ( ) ( ) e P = 0 =. 0! If there are 0 defectives, p = 0 00 and µ =.. Then ( ) ( ) 0.. e P = 0 =. 7 0! If there are 0 defectives, p = 0 00 and µ =.. Then ( ) ( ) 0.. e P = 0 =. 6 0!.0 The random variable, the number of families who select a restaurant because of its great food, has a binomial distribution with n = and p =.6. a Since p =.6 is not in Table, you must use the binomial formula to find b ( ) C ( ) ( ) 0 P = =.6. =.0 The probability that eactly three of the families select a restaurant because of its great food is ( ) C ( ) ( ) P = =.6. =.00 0 c P( ) P C = ( = 0) = (.6) (.) =.00 = The random variable, the number of offspring with Tay-Sachs disease, has a binomial distribution with n = and p =.. Use the binomial formula. 0 a P( ) C ( ) ( ) ( ) = =..7 =. =.06 b P( ) C ( ) ( ) ( )( ) c = =..7 =..7 =.7 Remember that the trials are independent. Hence, the occurrence of Tay-Sachs in the first two P third child develops Tay-Sachs =.. children has no affect on the third child, and ( ). The random variable, the number of college students who plan to move back home after college, has a binomial distribution with n = and p =.6. Use Table in Appendi I to find the necessary probabilities. P > 6 = P 6 =. =.66. a ( ) ( ) b P( ) =.07. c ( ) ( ) P = = P P( 9) = =.06. 9

17 . a The random variable, the number of tasters who pick the correct sample, has a binomial distribution with n= and, if there is no difference in the taste of the three samples, p = P(taster picks the correct sample) = b The probability that eactly one of the five tasters chooses the latest batch as different from the others is c P ( = ) = C =.9 The probability that at least one of the tasters chooses the latest batch as different from the others is 0 P ( ) = P ( = 0) = C 0 =.6. The random variable, the number of questionnaires that are filled out and returned, has a binomial distribution with n = 0 and p =.7. Use Table in Appendi I to find the necessary probabilities. a ( ) ( ) b P( ) P( ) c P( ) =.0. P = = P P( 9) =.0.07 =.0. = =. =.7.. Refer to Eercise.. a The average value of is µ = np = 0(.7) =. b The standard deviation of is σ = npq = 0(.7)(.) =. =.09. c The z-score corresponding to = is µ z = = =.9. σ.09 Since this z-score does not eceed in absolute value, we would not consider the value = to be an unusual observation..6 The random variable, the number of treated birds who are still infected with the parasite, has a binomial distribution with n =. If the diet supplement is ineffective, then the proportion of infected birds should still be p =., even after two weeks of treatment. a Using Table with n and p., we find = = ( ) P =.0. b If the treatment is effective, reducing the value of p to. p =, then ( ) P = The random variable has a Poisson distribution with µ =. Use Table in Appendi I or the Poisson formula to find the following probabilities. a b 0 e P ( = 0) = = e =. 0! 0 e e e P ( ) = + + 0!!! = = The random variable, the number of salesmen who will be involved in a serious accident during the coming year, has a binomial distribution with n = 0 and p =.0. To use the Poisson approimation, calculate µ = np = 0(.0) =. The probability that = is approimated as e P ( = ) = =.90!

18 .9 The random variable, the number of subjects who revert to their first learned method under stress, has a binomial distribution with n = 6 and p =.. The probability that at least five of the si subjects revert to their first learned method is P ( ) = P ( ) =. =.6.90 The random variable, the number of applicants who will actually enroll in the freshman class, has a binomial distribution with n = 60 and p =.9. Calculate µ = np = 60(.9) = and σ = npq = 60(.9)(.) =.06. Then approimately 9% of the values of should lie in the interval µ ± σ ± (.06) 0.7 to 6.. or between 0 and 6..9 The random variable, the number of California homeowners with earthquake insurance, has a binomial distribution with n = and p =.. a P ( ) = P ( = 0) =.06 =.79 b P ( ) = P ( ) =.9 =.06 c Calculate µ = np = (.) =. and σ = npq = (.)(.9) =.69. Then approimately 9% of the values of should lie in the interval µ ± σ. ± (.69). to.. or between 0 and..9 The random variable has a hypergeometric distribution with N =, M = and n=. Then P( = 0 ) = = =. and P ( = ) = = =. C 70 C 70.9 The random variable, the number of women who get fast food when they are too busy, has a binomial distribution with n = 0 and p =.6. a The average value of is µ = np = 0(.6) = 6. b The standard deviation of is σ = npq = 0(.6)(.6) =.0 =.. c The z-score corresponding to = 9 is µ 9 6 z = = =.7 σ. Since this value is not greater than in absolute value, it is not an etremely unusual observation. It is between and, however, so it is somewhat unusual..9 The student should follow the instructions for the Calculating Binomial Probabilities applet. When p =., the distribution is skewed right; when p =., the distribution is skewed left; and when p =., the distribution is symmetric about the mean..9 Use the Calculating Binomial Probabilities applet. The correct answers are given below. a P( < 6) = 6.0( ) = d P( < < 6 ) =.9 b P( = ) =.0 e P( 6) = c P( > ) = Define to be the number of successful operations. Then p P[ ] = success =. and n =. Use the Calculating Binomial Probabilities applet to calculate the necessary probabilities. 0 a P[ ] C ( ) ( ) ( ) = =.. =. =.77 b P[ ] C ( ) ( ) ( ) ( ) = =.. =.. =.096

19 0 c P[ ] p p C ( ) ( ) C ( ) ( ) < = (0) + () = = = It is known (part c, Eercise.96) that the probability of less than two successful operations is This is a very rare event, and the fact that it has occurred leads to one of two conclusions. Either the success rate is in fact 0% and a very rare occurrence has been observed, or the success rate is less than 0% and we are observing a likely event under the latter assumption. The second conclusion is more likely. We would probably conclude that the success rate for this team is less than 0% and would put little faith in the team..9 Define to be the number of failures observed among the four engines. Then p P[ ] q = p =. 99, with n =. a 0 P[ no failures] = P[ = 0 ] = C0 (.0) (.99) =.9606 b P[ no more than one failure] = P[ ] = p(0) + p() C ( ) ( ) C ( ) ( ) 0 0 = = =.999 = engine fails =.0 and.99 Define to be the number of young adults who prefer McDonald s. Then has a binomial distribution with n = 0 and p =.. Use the Calculating Binomial Probabilities applet. a P( ) b P( ) 6 0 = =.96 c If 0 prefer Burger King, then 60 prefer McDonalds, and vice versa. The probability is the same as that calculated in part b, since p =...0 Define to be the number of college seniors who want to take some time off after graduation. Then has a binomial distribution with n = 0 and p =.. a µ = np = 0(.) = 7 and σ = npq = 0(.)(.6) =. b Use the Calculating Binomial Probabilities applet to calculate P( ) =.007. This is a very unlikely event, happening less than one time in a thousand. c The z-score µ 7 z = = =.7 σ. indicates that = lies more than three standard deviations above the mean. This confirms the answer in part b. Case Study: A Mystery: Cancers Near a Reactor Let be the number of cancer cases in the state, and let µ be the average number of cancer cases statewide. If is 0% more than the average, then =. µ + µ or µ =. =.67 Since has a Poisson distribution with µ =.67, the standard deviation of is σ = µ =.67 =.9. c The z-score for the observed value of is µ.67 z = = =.9 σ.9 which is somewhat unlikely, under the assumption that the average is still µ =.67 for this area. It is more likely that there is a difference in the cancer rates for the area close to the reactor.