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Sample Chapter Chapter 5: Discrete Probability Distributions
Section 51: Probability Distributions
 The random variable is x, which is the number of girls in four births. The possible values of x are 0, 1, 2, 3, and
 The values of the random variable x are numerical.
 The random variable is discrete because the number of possible values is 5, and 5 is a finite number. The random variable is discrete if it has a finite number of values or a countable number of
 The table does describe a probability distribution because the three requirements are First, the variable
x is a numerical random variable and its values are associated with probabilities. Second,SP( x) = 0.063 + 0.250 + 0.375 + 0.250 + 0.063 = 1.001, which is not exactly 1 due to roundoff error, but is close enough to satisfy the requirement. Third, each of the probabilities is between 0 and 1 inclusive, as required.
 The probability of 0.136 is relevant; 56 is not significantly high because the probability of 56 or more girls is 0.136, which is not small, such as 0.05 or less. With random chance, it is likely that the outcome could be 56 or more girls.
 continuous random variable
 not a random variable
 discrete random variable
 continuous random variable
 discrete random variable
 Probability distribution with
 a. not a random variable
 continuous random variable
 discrete random variable
 not a random variable
 discrete random variable
m = 0 × 0.031+1×156 + 2 × 0.313 + 3× 313 + 4 × 0.156 + 5× 0.031 = 2.5o =TI Output for Exercise 7= 1.1
 Probability distribution (The sum of the probabilities is 1.001, but that is due to rounding errors.) with
m = 0 × 0.659 +1× 0.287 + 2 × 0.050 + 3× 004 + 4 × 0.001+ 5× 0 = 0.4 o =TI Output for Exercise 8= 0.6
 Not a probability distribution because the sum of the probabilities is 1, which is not 1 as required. Also, Ted clearly needs a new approach.
 Not a probability distribution because the responses are not values of a numerical random
 Probability distribution with
m = 0 × 0.091+1× 0.334 + 2 × 0.408 + 3× 0.166 = 1.6o = (0 – 1.6)^{2} × 0.091+ (1 1.6)^{2} × 0.334 + (2 – 1.6)^{2} × 0.408 + (3 – 1.6)^{2} × 0.166 = 0.9(The sum of the probabilities is 0.999, but that is due to rounding errors.)TI Output for Exercise 11
 Probability distribution with
m = 0 × 0.358 +1× 0.439 + 2 × 0.179 + 3× 0.024 = 0.9o =TI Output for Exercise 12 = 0.8
 This is not a probability distribution because the responses are not values of a numerical random
 This is not a probability distribution because the sum of the probabilities is 0.934 instead of 1 as required. The discrepancy between 0.934 and 1 is too large to attribute to rounding
 15.
m = 0 × 0.004 +1× 0.031+ 2 × 0.109 ++ 6 × 0.109 + 7 × 0.031+ 8× 0.004 = 4.0 girlso =TI Output for Exercise 15= 1.4 girls
 The lower limit is m – 2s = 0 – 2(1.4) = 1.2 girls. Because 1 girl is less than or equal to 1.2 girls, it is a significantly low number of girls.
 The upper limit is m + 2s = 0 + 2(1.4) = 6.8 girls. Because 6 girls is not greater than or equal to 6.8 girls, it is not a significantly high number of girls.
 b.
P( X = 7) = 0.031P( X ³ 7) = 0.031+ 0.004 = 0.035
 The probability from part (b), since it is the probability of the given or more extreme result.
 Yes, because the probability of 7 or more girls is 0.035, which is low (less than or equal to 0.05).
 b.
P( X = 6) = 0.109P( X ³ 6) = 0.109 + 0.031+ 0.004 = 0.144
 The result from part (b), since it is the probability of the given or more extreme result.
 No, because the probability of six or more girls is 0.144, which is not very low (less than or equal to 0.05).
 b.
P( X = 1) = 0.031P( X £ 1) = 0.004 + 0.031 = 0.035
 The result from part (b), since it is the probability of the given or more extreme result.
 Yes, because the probability of one or fewer girls is 0.035, which is low (less than or equal to 0.05). 21. m = 0 × 0.172 +1× 0.363 + 2 × 0.306 + 3× 0.129 + 4 × 0.027 + 5× 0.002 = 1.5 sleepwalkers
o =TI Output for Exercise 21= 1.0 sleepwalkers
 Significantly high numbers of sleepwalkers are greater than or equal to m + 2s =5 + 2(1.0) = 3.5 sleepwalkers. Because 4 sleepwalkers is greater than or equal to 3.5 sleepwalkers, 4 sleepwalkers is a significantly high number.
 Significantly high numbers of sleepwalkers are greater than or equal to m + 2s =5 + 2(1.0) = 3.5 sleepwalkers. Because 3 sleepwalkers is not greater than or equal to 3.5 sleepwalkers, 3 sleepwalkers is not a significantly high number.
 b.
P( X = 4) = 0.027P( X ³ 4) = 0.027 + 0.002 = 0.029
 The probability from part (b), since it is the probability of the given or more extreme result.
 Yes, because the probability of four or more sleepwalkers is 0.029, which is very low (less than or equal to 0.05)
 b.
P( X = 1) = 0.363P( X £ 1) = 0.172 + 0.363 = 0.535
 The probability from part (b), since it is the probability of the given or more extreme
 No, because the probability of one or fewer sleepwalkers is 0.535, which is not low (less than or equal to 0.05)
 a. 10 ×10 ×10 ×10 = 10, 000
 1 10, 000
 $5000 – $1 = $4999
 $1×1+ $5000 ×
1 10, 000= $0.50 = 50 ¢
 The $1 bet on the pass line in craps is better because its expected value of –1.4¢ is much greater than the expected value of –50¢ for the Ohio Pick 4
TI Output for Exercise 26 TI Output for Exercise 27
 a. 10 ×10 ×10 = 1000
 1 1000
 $500 – $1 = $499
 $1×1+ $500 ×
1 1000= $0.50 = 50 ¢
 Because both bets have the same expected value of –50¢, neither bet is better than the other.
$30 × 5 – $5× 33 = $0.39, or 39¢ 38 38
 The bet on the number 27 is better because its expected value of 26¢ is greater than the expected value of
39¢ for the other bet.
TI Output for Exercise 28 TI Output for Exercise 29
 Surviving the year: $161; Not surviving the year: $50, 000 – $226 = $99,839
 $161× 0.9986 + $99,839 × (1 0.9986) = $21
 Yes. The expected value for the insurance company is $21, which indicates that the company can expect to make an average of $21 for each such policy.
 a. –$226 and $49,774
 $226 × 0.9968 + $49, 774 × (1 0.9968) = $66
 Yes; the expected value for the insurance company is $66, which indicates that the company can expect to make an average of $66 for each such policy.
TI Output for Exercise 30
Section 52: Binomial Probability Distributions
 The given calculation assumes that the first two consumers are comfortable with the drones and the last three consumers are not comfortable with drones, but there are other arrangements consisting of two consumers who are comfortable and three who are not. The probabilities corresponding to those other arrangements should also be included in the result.
 n = 5, x = 2, p = 0.42, q = 0.58
 Because the 30 selections are made without replacement, they are dependent, not independent. Based on the 5% guideline for cumbersome calculations, the 30 selections can be treated as being independent. (The 30 selections constitute 3% of the population of 1009 responses, and 3% is not more than 5% of the population.) The probability can be found by using the binomial probability formula, but it would require application of that formula 21 times (or 10 times if we are clever), so it would be better to use
 The 0+ indicates that the probability is a very small positive value. (The actual value is 0.0000205.) The notation of 0+ does not indicate that the event is impossible; it indicates that the event is possible, but very unlikely.
 Not binomial; each of the weights has more than two possible
 binomial binomial
 Not binomial; each of the responses has more than two possible
 Not binomial; because the senators are selected without replacement, the selections are not independent. (The 5% guideline for cumbersome calculations cannot be applied because the 40 selected senators constitute 40% of the population of 100 senators, and that exceeds 5%.)
 Not binomial; because the senators are selected without replacement, they are not independent. (The 5% guideline for cumbersome calculations cannot be applied because the 10 selected senators constitute 10% of the population of 100 senators, and that exceeds 5%.). Also, the numbers of terms have more than two possible outcomes.
 Binomial; although the events are not independent, they can be treated as being independent by applying the 5% guideline. The sample size of 1019 is not more than 5% of the population of all
 Binomial; although the events are not independent, they can be treated as being independent by applying the 5% guideline. The sample size of 1000 is not more than 5% of the population of all
 a. 4 × 4 × 1 = 0.128
5 5 5
 {WWC, WCW, CWW}; Each has a probability of 0.128.
 0.128× 3 = 0.384
 a.
P(OOOD) = 0.4 × 0.4 × 0.4 × 0.6 = 0.0384
 {OOOD, OODO, ODOO, DOOO}; Each has a probability of 0.0384.
 0.0384 × 4 = 0.154
 8C7 × 0.2^{7} × 0.8^{1} = 0.0000819 (Table: 0+)
 8C4 × 0.2^{4} × 0.8^{4} + _{8}C5 × 0.2^{5} × 0.8^{3} + _{8}C6 × 0.2^{6} × 0.8^{2} + _{8}C7 × 0.2^{7} × 0.8^{1} + _{8}C8 × 0.2^{8} × 0.8^{0} = 0.0563 (Table: 0.056)
17._{8}C0 × 0.2^{0} × 0.8^{8} + _{8}C1 × 0.2^{1} × 0.8^{7} + _{8}C2 × 0.2^{2} × 0.8^{6} = 0.797(Table: 0.798) TI Output for Exercises 15, 17, and 19 TI Output for Exercises 16, 18, and 20
18.  _{8}C0 × 0.2^{0} × 0.8^{8} + _{8}C1 × 0.2^{1} × 0.8^{7} + _{8}C2 × 0.2^{2} × 0.8^{6} = 0.797  (Table: 0.798) 
19.  _{8}C0 × 0.2^{0} × 0.8^{8} = 0.168  20. 1 _{8}C0 × 0.2^{0} × 0.8^{8} = 0.832 (Table: 0.833) 
21.  _{8}C6 × 0.54^{6} × 0.46^{2} = 0.147TI Output for Exercises 21 and 23  22. 20 C15 × 0.54^{15} × 0.46^{5} = 0.0309TI Output for Exercises 22 and 24 
 10 C8 × 0.54^{8} × 0.46^{2} + _{10}C9× 0.54^{9} × 0.46^{1} + _{10}C10 × 0.54^{10} × 0.46^{0} = 0.0889
 10 C0 × 0.54^{0} × 0.46^{10} + _{10}C1 × 0.54^{1} × 0.46^{9} + _{10}C2× 0.54^{2} × 0.46^{8} = 0.00952
 90 C0 × 0.27^{0} × 0.73^{90} + _{90}C1 × 0.27^{1} × 0.73^{89} ++ _{90}C6 × 0.27^{6} × 0.73^{84} + _{90}C7 × 0.27^{7} × 0.73^{83} = 0.00000451; The
result of 7 minorities is significantly low. The probability shows that it is very highly unlikely that a process of random selection would result in 7 or fewer minorities. (The Supreme Court rejected the claim that the process was random.)
TI Output for Exercise 25 TI Output for Exercise 26
 20 C19× 79^{19} × 0.21^{1} + _{20}C20 × 0.79^{20} × 0.21^{0} = 0.0566; The probability of 19 or more adults requiring eyesight correction is not low (less than or equal to 0.05), so the result of 19 is not significantly high.
_{6}C5 × 0.20^{5} × 0.80^{1} = 0.002 (TI 83/84: 0.00154)
 6 C6 × 0.20^{6} × 0.80^{0} = 0+ (TI 83/84: 0.000064)
 0.002 + 0 = 0.002 (TI 83/84: 0.00160)
 Yes, the small probability from part (c) suggests that 5 is an unusually high number.
TI Output for Exercise 27 TI Output for Exercise 28
_{5}C0 × 0.20^{0} × 0.80^{5} = 0.328
 5C1 × 0.20^{1} × 0.80^{4} = 0.410
 0.328 + 0.410 = 0.738 (TI 83/84: 0.737)
 No, the probability from part (c) is not small, so 1 is not an unusually low number.
m = np = 36 × 0.5 = 18.0 girls, s = == 3.0 girls
 Values of 18.0 – 2(3.0) = 12.0 girls or fewer are significantly low, values of 18.0 + 2(3.0) = 24.0 girls or more are significantly high, and values between 12.0 girls and 24.0 girls are not significant.
 The result is significantly high because the result of 26 girls is greater than or equal to 24.0 girls. A result of 26 girls would suggest that the XSORT method is effective.
TI Output for Exercise 29 TI Output for Exercise 30
m = np = 16 × 0.5 = 8.0 girls, s = == 2.0 girls
 Values of 8.0 – 2(2.0) = 4.0 girls or fewer are significantly low, values of 8.0 + 2(2.0) = 12.0 girls or more are significantly high, and values between 4.0 girls and 12.0 girls are not significant.
 The result is not significant because the result of 11 girls is not greater than or equal to 12.0 girls. A result of 11 girls would not suggest that the XSORT method is effective.
m = np = 10 × 0.75 = 7.5 peas, s = == 1.4 peas
 Values of 7.5 – 2(1.4) = 4.7 peas or fewer are significantly low, values of 7.5 + 2(1.4) = 10.3 peas or more are significantly high, and values between 4.7 peas and 10.3 peas are not significant.
 The result is not significant because the result of 9 peas is not greater than or equal to 10.3 peas. TI Output for Exercise 31 TI Output for Exercise 32
m = np = 16 × 0.75 = 12.0 peas, s = == 1.7 peas
 Values of 12.0 – 2(1.7) = 8.6 peas or fewer are significantly low, values of 12.0 + 2(1.7) = 15.4 peas or more are significantly high, and values between 4.7 peas and 10.3 peas are not significant.
 The result is significant because the result of 7 peas is less than or equal to 8.6 peas.
 1 _{36}C0× 01^{0} × 0.99^{36} = 0.304; It is not unlikely for such a combined sample to test positive.
 1 _{8}C0× 1^{0} × 0.9^{8} = 0.570; It is likely for such a combined sample to test positive.
TI Output for Exercises 33 and 35 TI Output for Exercises 34 and 36
 40 C1× 03^{1} × 0.97^{39} + _{40}C0 × 0.03^{0} × 0.97^{40} = 0.662; The probability shows that about 2/3 of all shipments will be accepted. With about 1/3 of the shipments rejected, the supplier would be wise to improve quality.
 50 C2 × 0.02^{2} × 0.98^{48} + _{50}C1 × 0.02^{1} × 0.98^{49} + _{50}C0 × 0.02^{0} × 0.98^{50} = 0.922; About 92% of all shipments will be accepted. Almost all shipments will be accepted, and only 8% of the shipments will be rejected.
m = np = 100 × 0.16 = 16.0 M&Ms, s =np(1 p) =100 × 0.16 × 0.84 = 3.7 M&Ms; Values between 16.0 – 2(3.7) = 8.8 M&Ms and 16.0 + 2(3.7) = 23.4 M&Ms are not significant (8.7 and 23.3 if using unrounded values). 19 M&Ms lies between these limits, so it is not significant.
 The probability of exactly 19 green M&Ms is 100 C19 × 0.16^{19} × 0.84^{81} = 0.0736.
 The probability of 19 or more green M&Ms is 100 C19 × 0.16^{19} × 0.84^{81} ++ _{100}C100 × 0.16^{100} × 0.84^{0} = 0.242.
 The probability from part (c) is relevant. The result of 19 green M&Ms is not significantly high.
 The results do not provide strong evidence against the claim of 16% for green M&Ms.
TI Output for Exercise 37
m = np = 41× 0.5 = 20.5, s =np(1 p) =41× 0.5× 0.5 = 3.2; Values between 20.5 – 2(3.2) = 14.1 and 20.5 + 2(3.2) = 26.9 are not significant. The value of 40 is greater than or equal to 26.9, so it is significant.
 The probability of exactly 40 top lines for Democrats is 41C40 ×5^{40} × 0.5^{1}, which is approximately zero when rounded.
 The probability of 40or more top lines for Democrats is 41C40 ×5^{40} × 0.5^{1} + _{41}C41 × 0.5^{41} × 0.5^{0}, which is approximately zero when rounded.
 The probability from part (c) is The result of 40 top lines for Democrats is significantly high.
 The results suggest that the clerk did not assign ballot lines
TI Output for Exercise 38
m = np = 611× 0.43 = 262.7 votes, s =np(1 p) =611× 0.43× 0.57 = 12.2 votes; Values between 262.7 – 2(12.2) = 238.3 votes and 262.7 + 2(12.2) = 287.1 votes are not significant. The value of 308 votes is greater than or equal to 128.1, so it is significant.
 The probability of exactly 308 voters is 611C308 × 0.43^{308} × 0.57^{303} = 0.0000369.
 The probability of 308 or more voters is 611C308 × 0.43^{308} × 0.57^{303} ++ _{611}C611 × 0.43^{0} × 0.57^{611} = 0.000136.
 The probability from part (c) is relevant. The value of 308 votes is significantly high.
 The results suggest that the surveyed voters either lied or had defective memory of how they voted.
TI Output for Exercise 39
m = np = 580 × 0.25 = 145.0 yellow peas, s =np(1 p) =580 × 0.25× 0.75 = 10.4 yellow peas; Values between 145.0 – 2(10.4) = 124.2 yellow peas and 145.0 + 2(10.4) = 165.8 yellow peas are not significant (124.1 and 165.9 if using unrounded values). 152 yellow peas lies between these limits, so it is not significant.
 The probability of exactly 152 yellow peas is 580 C152 × 0.25^{152} × 0.75^{428} = 0.0301.
 The probability of 152 or more yellow peas is 580 C152 × 0.25^{152} × 0.75^{428} ++ _{580}C580 × 0.25^{580} × 0.75^{0}
= 0.265.
 The probability from part (c) is relevant. The value of 152 yellow peas is not significantly high.
 The results do not provide strong evidence against Mendel’s claim of 25% for yellow peas.
TI Output for Exercise 40
 41.
P(5) = 0.06 (1 0.06)^{4} = 0.0468
 42.
15!æ 18 ö^{7} æ 18 ö^{6} æ 2 ö^{2}× ç ÷ × ç ÷ × ç ÷ = 0.0302 7!6!2! è 38 ø è 38 ø è 38 ø
TI Output for Exercise 41 TI Output for Exercise 42
 43.
P(4) =6! ×43!¸ (6 + 43)!= 0.132 (6 – 2)!2! (43 – 6 + 2)!(6 – 2)! (6 + 43 – 6)!6!
TI Output for Exercise 43
Section 53: Poisson Probability Distributions
 m = 535 576 =929, which is the mean number of hits per region. x = 2,
because we want the probability that a randomly selected region had exactly 2 hits, and e » 2.71828 which is a constant used in all applications of Formula 59.
 The mean is m = 153 64 =4 tornadoes, the standard deviation is s = = 1.5 tornadoes, and the variance is s ^{2} = 2.4 tornadoes^{2}.
 The possible values of x are 0, 1, 2, … (with no upper bound), so x is a discrete random variable. It is not possible to have x =3 calls in a day
 P(0) represents the probability of no occurrences of an event during the relevant interval. If x = 0,
P(0) = e^{–}m .
6.15 × e6.1

P(5) 0.1585!
 In 55 years, the expected number of years with 5 hurricanes is 55× 0.158 = 8.7.
 The expected value of 8.7 years is close to the actual value of 8 years, so the Poisson distribution works well here.
6.10 × e6.1

P(0) 0.002240!
 In 55 years, the expected number of years with 5 hurricanes is 55× 0.00224 = 0.1.
 The expected value of 0.1 years is close to the actual value of 0 years, so the Poisson distribution works well here.
TI Output for Exercises 5 and 7 TI Output for Exercises 6 and 8

6.17 × e6.1P(7) 0.1407!
 In 55 years, the expected number of years with 5 hurricanes is 55× 0.140 = 7.7.
 The expected value of 7.7 years is close to the actual value of 7 years, so the Poisson distribution works well here.
6.14 × e6.1

P(4) 0.1294!
 In 55 years, the expected number of years with 5 hurricanes is 55× 0.129 = 7.1.
 The expected value of 7.1 years is not very close to the actual value of 10 years, so the Poisson distribution does not work so well here.
422111.615 × e11.6
 m = =6 births, P(15) = = 0.0649 (0.0643 if using the unrounded mean.) There is less
365 15!than a 7% chance of getting exactly 15 births on any given day. 3330.90 × e0.9
 m = =9 murders, P(0) = = 0.407 (0.402 if using the unrounded mean.) There should be many
365 0!days (roughly 40%) with no murders.
TI Output for Exercises 9 and 11 TI Output for Exercise 10
b.m = 22, 713 = 62.2365

62.250 × e62.2P(50) 0.015550!
 12.
m = 196 = 0.7280

9.80 × e0.7
 P(0) 0.497
0!

9.81 × e0.7
 P(1) 0.348
1!

9.82 × e0.7
 P(2) 0.122
2!

9.83 × e0.7
 P(3) 0.0284
3!

9.84 × e0.7
 P(4) 0.00497; The expected
4!TI Output for Exercise 12 frequencies of 139, 97, 34, 8 , and 1.4 compare reasonably well to the actual frequencies, so the Poisson distribution provides good results.
0.9292 × e0.929

P(2) 0.1702!
 The expected number of regions with exactly 2 hits is between 97.9 and 98.2, depending on
 The expected number of regions with 2 hits is close to 93, which is the actual number of regions with 2
TI Output for Exercise13 TI Output for Exercise 14
b.m = 12, 429 × 0.000011 = 0.1367

0.1370 × e0.137P(0) 0.8720!0.1371 × e0.137

and P(1) 0.119; So the probability of 0 or 1 is1! 0.872 + 0.119 = 0.991.
 1 0.991 = 0.009
 No, the probability of more than one case is extremely small, so the probability of getting as many as four cases is even
33, 56111.30 × e11.3
 m = =3 fatalities,1 P(0) = 1 = 0.9999876 or 0.9999877 if using the unrounded mean.
2969 0!There is a very high chance (“almost certain”) that at least one fatality will occur.
TI Output for Exercises 15 and 17 TI Output for Exercise 16 1810.50 × e0.5
 m = =5 fatalities, 1 P(0) = 1 = 0.393 or 0.391 if using the unrounded mean.
365 0!
 The Poisson distribution approximation is valid since n = 5200 ³ 100 and m = np =
5200 292, 201, 338 = 0.0000178 £ 10. The probability of winning at least one time is 1 P(0) = 10.00001780 × e0.0000178 0! = 0.0000178, so it is highly unlikely that at least one jackpot win will occur in 50 years.
Chapter Quick Quiz
 No, the sum of the probabilities is 4 3, or 1.333, which is greater than 1.
 m = 80 × 0.2 = 16.0; s = = 3.6
 The values are parameters because they represent the mean and standard deviation for the population of all who make random guesses for the 80 questions, not a sample of actual
Review Exercises 93
 No, (Using the range rule of thumb, the limit separating significantly high values is m + 2s = 0 + 2(3.6)
= 23.2, but 20 is not greater than or equal to 23.2. Using probabilities, the probability of 20 or more correct answers is 0.163, which is not low.)TI Output for Exercises 2 and 4 TI Output for Exercises 5 and 6
 Yes, (Using the range rule of thumb, the limit separating significantly low values is m – 2s =0 – 2(3.6)
= 8.8, and 8 is less than 8.8. Using probabilities, the probability of 8 or fewer correct answers is 0.0131, which is low.)
 This is probability distribution because the three requirements are satisfied. First, the variable x is a numerical random variable and its values are associated with Second, SP( x) = 0 + 0.006 + 0.051+ 0.205 +
0.409 + 0.328 = 0.999, which is not exactly 1 due to roundoff error, but is close enough to satisfy the requirement. Third, each of the probabilities is between 0 and 1 inclusive, as required.
 m = 0 × 0 +1× 0.006 + 2 × 0.051+ 3× 0.205 + 4 × 0.409 + 5× 0.328 = 4.0 flights
TI Output for Exercise 7
 s 2 = (0.9 flight)2 = 0.8 flight2
 0+ indicates that the probability is a very small positive number. It does not indicate that it is impossible for none of the five flights to arrive on time.
 P( X < 3) = P( X £ 2) = 0 + 0.006 + 0.051 = 0.057
Review Exercises
 P( X = 3) = _{5}C3× 0.74^{3} × 0.26^{2} = 0.247
 P( X ³ 1) = 1 P( X = 0) = 1 _{5}C0× 0.74^{0} × 0.26^{5} = 0.999; No; the five friends are not randomly selected from the population of adults. Also, the fact that they are vacationing together suggests that their financial situations are more likely to include credit cards.
 m = 5× 0.74 = 3.7, s = = 1.0
 No, the limit separating significantly high values is m + 2s =7 + 2(1.0) = 5.7, but 5 is not greater than or
equal to 5.7. Also, the probability that all five adults have credit cards is 5C5 × 0.74^{5} × 0.26^{0} = 0.222, which is not low (less than or equal to 0.05).
 Yes, the limit separating significantly low values is m – 2s =7 – 2(1.0) = 1.7, and 1 is less than or equal to
1.7. Also, the probability of one or fewer adults having a credit card is 5C0 × 0.74^{0} × 0.26^{5} + _{5}C1 × 0.74^{1} × 0.26^{4}= 0.0181, which is low (less than or equal to 0.05).
TI Output for Exercise 5 TI Output for Exercise 7
 This is not a probability distribution because the responses are not values of a numerical random variable. 7. This is not a probability distribution because SP( x) = 0016 + 0.0250 + 0.1432 + 0.3892 + 0.4096 = 0.9686,
instead of 1 as required. The discrepancy between 0.9686 and 1 is too large to attribute to rounding errors.
 This is a probability distribution (The sum of the probabilities is 0.999, but that is due to rounding errors.) with
m = 0 × 0 +1× 0.003 + 2 × 0.025 + 3× 0.111+ 4 × 0.279 + 5× 0.373 + 6 × 0.208 = 4.6 people and o =TI Output for Exercise 8= 1.0 people.
 a. 784 × 0.301 = 236.0 checks
 m = 784 × 0.301 = 236.0, checks s = 784 × 0.301× 0.699 = 12.8 checks
 The limit separating significantly low values is m – 2s = 236.0 – 2(12.8) = 210.4.
 Yes, because 0 is less than or equal to 210.4 checks.
TI Output for Exercise 9 TI Output for Exercise 10
 a. b.
m = 7 365 = 0.0192

0.01920 × e0.0192P( X 0) 0.9810! 0.01920 × e0.0192 0.01921 × e0.0192
 P( X > 1) = P( X ³ 2) = 1 P( X £ 1) = 1 – = 0.000182
0! 1!
 No, because the event is so rare. (But it is possible that more than one death occurs in a car crash or some other such event, so it might be wise to consider a contingency plan.)
Cumulative Review Exercises
 a. The mean is x = 0 +0 +1+ 2 + 8 +17 + 21+ 28 = 9.6 moons.
8
 The median is 2 +8 =0 moons.
2
 The mode is 0
 The range is 28 – 0 =0 moons.

(0 – 9.6)^{2} + (0 – 9.6)^{2} ++ (21 9.6)^{2} + (28 – 9.6)^{2}
 The standard deviation is s 0 moons.
8 – 1_{2} (0 – 9.6)^{2} + (0 – 9.6)^{2} ++ (21 9.6)^{2} + (28 – 9.6)^{2} _{2}
 The variance is s
= = 120.3 moons .8 – 1
 The minimum is 9.6 – 2 ×0 = 12.4 moons and the maximum is 9.6 + 2 ×11.0 = 31.6 moons.
 No, because none of the planets have a number of moons less than or equal to m – 2s =6 – 2(11.0) = 12.4
moons (which is impossible, anyway) and none of the planets have a number of moons equal to or greater thanm + 2s = 9.6 + 2(11.0) = 31.6 moons.
 ratio
 discrete
 a. 1 × 1 × 1 =
1 = 0.001TI Output for Exercise 2 10 10 10 1000
 365× 0.001 = 0.365
0.3651 × e0.365
 P(1) = = 0.254 1!
 1× 0.999 + 499 × 0.001 = 0.50, or 50¢.
 Refer to the following table.
Challenge Upheld with Overturned Call  Challenge Rejected with Overturned Call  Total  
Challenges by Men  152  412  564 
Challenges by women  79  236  315 
Total  231  648  879 
 231 =
77 = 0.263
 564 +231 152 = 643 = 0.732
879 293879 879 879 879
 79 231
= 0.342
 152 = 0.658
231
 231 × 230 = 0.0688
879 878 TI Output for Exercise 3
 a. 0.73× 347 = 253
 The sample is the 347 human resource professionals who were surveyed. The population is all human resource
 73% is a statistic because it is a measure based on a sample, not the entire
 No vertical scale is shown, but a comparison of the numbers shows that 7,066,000 is roughly 1.2 times the number 6,000,000. However, the graph makes it appear that the goal of 7,066,000 people is roughly 3 times the number of people enrolled. The graph is misleading in the sense that it creates the false impression that actual enrollments are far below the goal, which is not the case. Fox News apologized for their graph and provided a corrected
 b.
P( X = 5) = _{8}C5 × 0.7^{5} × 0.3^{3} = 0.254P( X ³ 7) = _{8}C7 × 0.7^{7} × 0.3^{1} + _{8}C8 × 0.7^{8} × 0.3^{0} = 0.255 (Table: 0.256)
 m = 8× 0.7 = 5.6 adults, s = = 1.3 adults
 (Using the range rule of thumb, the limit separating significantly low values is
m – 2s = 5.6 – 2(1.3) = 3.0, and 1 is less than 3. Using probabilities, the probability of 1 or fewer peoplewashing their hands is 8C0 × 0.7^{0} × 0.3^{8} + _{8}C1 × 0.7^{1} × 0.3^{7} = 0.00129, (Table: 0.001) which is low, such as less than 0.05.
TI Output for Exercise 6