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Chapter 6
Discrete Probability Distributions
Distribution, mean and standard deviation of discrete random variables are described, first in general, then for the binomial and Poisson special cases.
6.1 Discrete Random Variables
A random variable, denoted by a capital letter such as X, is a “rule” which assigns a number to each outcome in sample space of a probability experiment^1. A random variable is discrete if outcomes assigned to finite or countably infinite real values.
Exercise 6.1 (Discrete Random Variables)
Properties of discrete probability distribution of random variable X, value x.
∑ P (x) = 1, 0 ≤ P (x) ≤ 1
- expected value, mean: μX =
∑ [x · P (x)]
∑ [(x − μ)^2 P (x)] =
∑ [x^2 · P (x)] − μ^2 X
√ σ X^2
- Discrete or Continuous?
(a) discrete / continuous. Number of seizures in a year. (b) discrete / continuous. Waiting time at Burger King. (c) discrete / continuous. Temperature in Michigan City. (d) discrete / continuous. Number of bikes on bike rack. (e) discrete / continuous. Number of heads in three coin tosses. (f) discrete / continuous. Number of pips in roll of dice. (^1) More technically, a random variable is a function mapping from sample space to real line.
100 Chapter 6. Discrete Probability Distributions (Lecture Notes 6)
(g) discrete / continuous. Patient’s ages.
- Sample distribution versus probability distribution: seizures.
(a) Sample distribution. If data from 100 epileptic people sampled at random in one year was number number seizures people 0 17 2 21 4 18 6 11 8 16 10 17
observed average number of seizures would be
¯x =
which is equal to (circle one) 4. 32 / 4. 78 / 5. 50 / 5. 75. (StatCrunch: Relabel var1 as seizures, var2 as number. Type data into seizures and number columns. Stat, Summary Stats, Grouped/Binned data, Bins in: seizures, Counts in: number, Statistics: Mean, Std. dev., Compute. Notice sample ¯x = 4.78.) Sample average ¯x is a (circle one) parameter / statistic.
Observed standard deviation in number of seizures,
s =
√ 17(0 − 4 .78)^2 + · · · + 17(10 − 4 .78)^2 100 − 1
=
√ (0 − 4 .78)^2 ·
+ · · · + (10 − 4 .78)^2 ·
(circle one) 3. 32 / 3. 49 / 3. 50 / 3. 75 , (StatCrunch: Notice sample s ≈ 3 .49.) Sample SD s is also a (circle one) parameter / statistic. (b) Probability distribution. Table of probability distribution:
102 Chapter 6. Discrete Probability Distributions (Lecture Notes 6)
ii. Function (b).
P (X = x) =
{
- 18 , if x = 4
- 11 , if x = 6 iii. Function (c).
P (X = x) =
17 , if x = 0
21 , if x = 2
18 , if x = 4
11 , if x = 6
16 , if x = 8
17 , if x = 10
Probability distribution, mean μX and SD σX : number of seizures, X
number seizures, x P (x) 0 0. 17 2 0. 21 4 0. 18 6 0. 11 8 0. 16 10 0. 17
(a) Various probabilities associated with number of seizures. i. Chance a person has 8 epileptic seizures is P (8) = P (X = 8) = (circle one) 0. 14 / 0. 15 / 0. 16 / 0. 17. ii. Chance a person has at most 4 seizures is P (X ≤ 4) = P (0) + P (2) + P (4) = (circle one) 0. 17 / 0. 21 / 0. 56 / 0. 67. iii. Chance a person has at least 4 seizures is P (X ≥ 4) = P (4) + P (6) + P (8) + P (10) = 1 − P (X ≤ 3) = (circle one) 0. 21 / 0. 38 / 0. 56 / 0. 62. iv. P (0) + P (2) + P (4) + P (6) + P (8) + P (10) = 0. 97 / 0. 98 / 1. v. P (2.1) = (circle one) 0 / 0. 21 / 0. 56 / 0. 67. (b) Mean (expected value) of number of seizures.
μX =
∑ [x · P (x)] = 0 · P (0) + 2 · P (2) + 4 · P (4) + 6 · P (6) + 8 · P (8) + 10 · P (10) = 0(0.17) + 2(0.21) + 4(0.18) + 6(0.11) + 8(0.16) + 10(0.17) =
(circle one) 4. 32 / 4. 78 / 5. 50 / 5. 75. (Stat, Calculators, Custom, Values in: seizures, Weights in: P(x), Okay. Notice μX = Mean: 4.78.)
Understanding mean (expected value): point of balance.
Section 1. Discrete Random Variables (Lecture Notes 6) 103
(a)
number seizures, x (b)
number seizures, x (c)
number seizures, x
P(X = x)
(^0 2 4 6 )
P(X = x)
(^0 2 4 6 )
P(X = x)
(^0 2 4 6 )
Figure 6.2 (Mean number of seizures: point of balance.) Mean balances “weight” of probability in graph (a) / (b) / (c). In other words, mean (expected value) close to (circle one) 1 / 5 / 9. (c) Standard deviation in number of seizures.
σX =
√∑ [(x − μX )^2 P (x)]
=
√ (0 − 4 .78)^2 (0.17) + (2 − 4 .78)^2 (0.21) + · · · + (10 − 4 .78)^2 (0.17) ≈
(circle one) 3. 47 / 4. 11 / 5. 07 / 6. 25. (Stat, Calculators, Custom, Values in: seizures, Weights in: P(x), Okay. Notice σX = Std, Dev.: 3.47.)
Understanding standard deviation: dispersion (spread).
(a) seizure distribution
P(X = x)
(^0 2 4 6 )
(b) another distribution (c) and another distribution
P(X = x)
0 2 4 6 8 10
P(X = x)
(^0 2 4 6 )
4.78 + 3.47_ 4.78 + 2.47_ 4.78 + 1.47_
number seizures, x (^) number seizures, x number seizures, x
Figure 6.3 (SD in number of seizures: dispersion (spread).) Standard deviation measures dispersion of a probability distribution. Most dispersed distribution occurs in (a) / (b) / (c). We expect to see about 4.78 “±” 3.47 seizures according to seizure probability distribution.
Variance in number of seizures is σ^2 ≈ 3. 472 ≈ (circle one) 10. 02 / 11. 11 / 12. 07 / 13. 25. (Stat, Calculators, Custom, Values in: seizures, Weights in: P(x), choose Variance, Okay.)
Section 1. Discrete Random Variables (Lecture Notes 6) 105
i. Chance bike rack has 8 bicycles is P (8) = (circle one) 15 / 25 / 35 / 45. ii. Chance bike rack has at most 6 bicycles is P (X ≤ 6) = P (5) + P (6) = (circle one) 15 / 25 / 35 / 45. iii. Chance bike rack has at least 6 bicycles is P (X ≥ 6) = P (6) + P (7) + P (8) + P (9) = 1 − P (X ≤ 5) = (circle one) 15 / 25 / 35 / 45. iv. Chance bike rack has more than 6 bicycles is P (X > 6) = P (X ≥ 7) = P (7) + P (8) + P (9) = 1 − P (X ≤ 6) = (circle one) 15 / 25 / 35 / 45. (d) Mean (expected value) of number of bikes.
μX =
∑ [x · P (x)]
= 5 ×
+ 6×
+ 7×
+ 8×
+ 9×
(circle one) 5 / 6 / 7 / 8. (StatCrunch: Blank data table. Relabel var1 bikes, var2 P(x). Data, Data save 6.1.3 bike distribution. Stat, Calculators, Custom, Values in: bikes, Weights in: P(x), Okay. Notice Mean: 7 or notice 7 is balance point of probability histogram.) (e) Standard deviation in number of bikes.
σX =
√∑ [(x − μX )^2 P (x)]
=
√ (5 − 7)^2 (0.2) + (6 − 7)^2 (0.2) + · · · + (9 − 7)^2 (0.2) ≈
(circle one) 1. 41 / 2. 41 / 3. 07 / 4. 25. (StatCrunch: Notice Std. Dev.: 1.41.) In other words, we expect to see about 7 “±” 1.4 bikes on bike rack.
- Probability distribution, mean μX and SD σX : roulette payoff, X Roulette table has 38 numbers: numbers are 1 to 36, 0 and 00. A ball is spun on a roulette wheel. After a time, ball drops into one of 38 slots which correspond to 38 numbers on roulette table.
(a) Betting on even. Let random variable X be payoff from $1 bet on even: $1 lost if ball drops on odd or 0 or 00, $1 won (added to $1 bet) if even.
payoff, x P (x) -$1 (^2038) $1 (^1838)
106 Chapter 6. Discrete Probability Distributions (Lecture Notes 6)
Mean is μX = − 1 × 2038 + 1 × 1838 = − 382 (≈ − 0 .05) and so
σX =
√( − 1 −
( −
)) 2 20 38
( 1 −
( −
)) 2 18 38
(circle one) 0. 051 / 0. 999 / 1. 573 / 2. 251 (StatCrunch: Blank data table. Relabel var1 payoff, var2 frequency, var3 P(x). Since StatCrunch deals with fractions awkwardly, first type -1, 1 into payoff, and 20, 18 into frequency, then, to derive probabilities, Data, Compute expression, Expression: frequency/38, New column name: P(x), Compute. Data, Data save 6.1.4 roulette even distribution. Stat, Calculators, Custom, Values in: payoff, Weights in: P(x), Okay. Notice Std. Dev.: 0.999.)
We expect to lose a nickel “±” a dollar betting $1 on even.
Also, σ X^2 ≈ 0. 9992 ≈ (circle one) 0. 997 / 0. 998 / 0. 999 / 1. 000. (Stat, Calculators, Custom, Values in: payoff, Weights in: P(x), choose Variance, Okay.) (b) Betting on a section. Let random variable Y be payoff from a $1 bet on a section (with 12 numbers): $1 lost if ball drops on one of 24 numbers not in section or 0 or 00, $2 won (added to $1 bet) if number in section.
payoff, x P (x) -$1 (^2638) $2 (^1238)
Mean is μY = − 1 × 2638 + 2 × 1238 = − 382 (≈ − 0 .05) and so
σY =
√( − 1 −
( −
( 2 −
( −
(circle one) 0. 05 / 0. 47 / 1. 39 / 2. 25 (StatCrunch: Blank data table. Relabel var1 payoff, var2 frequency, var3 P(x). Since StatCrunch deals with fractions awkwardly, first type -1, 2 into payoff, and 26, 12 into frequency, then, to derive probabilities, Data, Compute expression, Expression: frequency/38, New column name: P(x), Compute. Data, Data save 6.1.5 roulette section distribution. Stat, Calculators, Custom, Values in: payoff, Weights in: P(x), Okay. Notice Std. Dev.: 1.39.)
We expect to lose a nickel “±” $1.39 betting $1 on a section.
Also, σ Y^2 ≈ 1. 392 = (circle one) 1. 945 / 1. 946 / 1. 947 / 1. 948.
What is expected on average when betting $100 on a section? μY = − 100 × 2638 + 200 × 1238 ≈ (circle one) − 5. 26 / − 9. 99 / − 15. 73
108 Chapter 6. Discrete Probability Distributions (Lecture Notes 6)
(a)
P(X = x)
(^0 1 2 3 ) number cases won, x
5 6 7 8 9
(b)
P(X = x)
0 1 2 3 4 10 number cases won, x
5 6 7 8 9
(c)
P(X = x)
0 1 2 3 4 10 number cases won, x
5 6 7 8 9
Figure 6.5 (Probability histogram of number of wins.) Probability histogram, number of wins: (choose one) (a) / (b) / (c). (b) Probability function, number of wins. Choose two! i. Function (a).
P (X = x) =
x 46
, x = 0, 1 , 2 ,... , 10.
(Type 0,1,2... , 10 into L 1. Define L 2 = L 1 ÷ 46 ENTER. Probabilities in L 2 equal to P (x)?) ii. Function (b).
P (X = x) =
- 006 , if x = 0,
- 040 , if x = 1,
- 121 , if x = 2, .. .
- 002 , if x = 9,
- 000 , if x = 10.
iii. Function (c).
P (X = x) = 10 Cx 0. 4 x(1 − 0 .4)^10 −x, x = 0, 1 , 2... , n
(StatCrunch: Blank data table. Relabel var1 case, var2 P(x). Type number of cases, 0, 1, ..., 10 into case column. To derive probabilities, Data, Compute expression, Expression: dbi- nom(case,10,0.4), New column name: P(x), Compute. Data, Data save 6.2.1 binomial lawyer distribution.) (c) Various probabilities associated with number of wins. i. Chance lawyer wins 8 cases is
P (8) = 10 C 8 p^8 (1−p)^10 −^8 =
(0.4)^8 (1− 0 .4)^2 = 45· 0. 480. 62 ≈
(circle one) 0. 006 / 0. 011 / 0. 040 / 0. 121. (StatCrunch: Stat, Calculators, Binomial, n: 10, p: 0.4, Prob(X = 8) (not Prob(X ≤ 8)!) Compute. Notice Prob(X = 8) = 0.011.)
Section 2. The Binomial Probability Distribution (Lecture Notes 6) 109
ii. Chance lawyer has at most 6 wins is
P (X ≤ 6) = P (0) + P (1) + P (2) + P (3) + P (4) + P (5) + P (6) = 10 C 0 p^0 (1 − p)^10 −^0 + 10 C 1 p^1 (1 − p)^10 −^1 + · · · + 10 C 6 p^6 (1 − p)^10 −^6
=
(0.4)^0 (1 − 0 .4)^10 + · · · +
(0.4)^6 (1 − 0 .4)^4
(circle one) 0. 834 / 0. 934 / 0. 945 / 0. 993. (Long way: 1 × 0. 4 ˆ(0) × 0. 6 ˆ(10) + 10 × 0. 4 ˆ(1) × 0. 6 ˆ(9) + · · · + 210 × 0. 4 ˆ(6) × 0. 6 ˆ(4) OR Short way: Stat, Calculators, Binomial, n: 10, p: 0.4, Prob(X ≤ 6) Compute. ) iii. Chance lawyer has less than 6 wins is P (X < 6) = P (X ≤ 5) ≈ (circle one) 0. 834 / 0. 934 / 0. 945 / 0. 993. (StatCrunch: Stat, Calculators, Binomial, n: 10, p: 0.4, Prob(X < 6) Compute.) (d) Mean (expected value) of number of wins. long way:
μX =
∑ [x · P (x)] = 0 × 0 .006 + 1× 0 .040 + · · · + 10× 0. 000 ≈
(circle one) 3 / 4 / 5 / 6. (StatCrunch: Stat, Calculators, Custom, Values in: case, Weights in: P(x), Compute. Notice, Mean: 4)
short way: μX = np = 10 × 0 .4 = (circle one) 3 / 4 / 5 / 6. (e) Standard deviation in number of wins. long way:
σX =
√∑ [(x − μX )^2 P (x)]
=
√ (0 − 4)^2 (0.006) + (1 − 4)^2 (0.040) + · · · + (10 − 4)^2 (0.000) ≈
(circle one) 1. 23 / 1. 44 / 1. 55 / 1. 76. (Stat, Calculators, Custom, Values in: case, Weights in: P(x), Compute. Notice, Std. Dev.: 1.59)
short way: σX =
√ np(1 − p) =
√ 10(0.4)(1 − 0 .4) ≈ (circle one) 1. 23 / 1. 44 / 1. 55 / 1. 76.
We expect lawyer to win about 4 “±” 1.6 of 10 trials.
Section 2. The Binomial Probability Distribution (Lecture Notes 6) 111
(b) Chance seven widgets defective P (7) = 14 C 7 × (0.21)^7 × (1 − 0 .21)^14 −^7 ≈ 0. 005 / 0. 012 / 0. 040. (StatCrunch: Stat, Calculators, Binomial, n: 14, p: 0.21, Prob(X = 7) Compute.) (c) Chance at most ten widgets defective P (X ≤ 10) = P (0) + P (1) + · · · + P (10) ≈ 0. 995 / 0. 997 / 0. 999. (StatCrunch: Stat, Calculators, Binomial, n: 14, p: 0.21, Prob(X ≤ 10) Compute.) (d) Chance at least ten widgets defective P (X ≥ 10) = P (10) + P (11) + P (12) + P (13) + P (14) = 1 − P (X ≤ 9) ≈ (circle one) 0. 000072 / 0. 00072 / 0. 0072. (StatCrunch: Stat, Calculators, Binomial, n: 14, p: 0.21, Prob(X ≥ 10) Compute.) (e) Chance between 7 and 10 widgets defective, inclusive P (7 ≤ X ≤ 10) = P (7)+P (8)+P (9)+P (10) = P (X ≤ 10)−P (X ≤ 6) ≈ (circle one) 0. 005 / 0. 015 / 0. 034. (Stat, Calculators, Binomial, Between, n: 14, p: 0.21, Prob(7 ≤ X ≤ 10) Compute.)
Since P (7 ≤ X ≤ 10) ≈ 0. 015 < 0 .05, it is / is not unusual to have 7 and 10 widgets defective, inclusive. (f) Expected number of defectives μX = np = 14(0.21) = (circle one) 2. 44 / 2. 51 / 2. 94. (g) Standard deviation in number of defectives: σX =
√ np(1 − p) ≈ (circle one) 1. 52 / 1. 63 / 1. 76. So, expect 2.94 “±” 1.52 defectives. (h) Probability histograms, for different chance each widget defective, p. Identify different p for different probability histograms.
P(X = x)
0 1 2 3 4 10 number defective widgets, x
5 6 7 8 9
11 12 1314
P(X = x)
0 1 2 3 4 10 number defective widgets, x
5 6 7 8 9
11 12 1314
P(X = x)
0 1 2 3 4 10 number defective widgets, x
5 6 7 8 9
11 12 1314
p = (choose one) 0.21 / 0.52 / 0.83 p = (choose one) 0.21 / 0.52 / 0.83 p = (choose one) 0.21 / 0.52 / 0.
Figure 6.6 (Probability histograms: number of defective, different p.)
Notice, probability histograms for p = 0.21, p = 0.52 and p = 0.83 are right-skewed, (more or less) symmetric and left-skewed, respectively.
112 Chapter 6. Discrete Probability Distributions (Lecture Notes 6)
- Binomial: number of correct multiple choice answers. On a multiple–choice exam with 4 possible choices for each of 5 questions, what is probability a student gets 3 or more correct answers just by guessing?
(a) Since there are five questions, n = (circle one) 3 / 4 / 5 (b) Student wants 3 or more correct answers, x = (circle one) 3 / 3 , 4 / 3 , 4 , 5 (c) Each question has 4 possible choices and student is choosing at random, p = (circle one) 13 / 14 / (^15) (d) Chance student gets at least 3 correct answers just by guessing is P (X ≥ 3) = P (3) + P (4) + P (5) = 1 − P (X ≤ 2) ≈ (choose one) 0. 097 / 0. 104 / 0. 112 (StatCrunch: Stat, Calculators, Binomial, n: 5, p: 0.25, Prob(X ≥ 3) Compute.)
Since P (X ≥ 3) ≈ 0. 104 > 0 .05, it is / is not unusual to get at least 3 correct answers just by guessing. (e) Expected number of correct answers μX = np = 5 × 14 = (circle one) 14 / 34 / 54 = 1.25. (f) Standard deviation in number of correct answers: σX =
√ np(1 − p) =
√ 5 × (^14)
( 1 − (^14)
) ≈ (circle one) 0. 52 / 0. 97 / 1. 06. So, expect 1.25 “±” 0.97 correct answers. (g) Probability histogram, large n. As n increases; specifically, for np(1 − p) ≥ 10, any binomial is symmetric, in fact, bell-shaped enough so empirical rule^2 can be used. Since
np(1 − p) = 5 × 0. 25 × (1 − 0 .25) = 0. 9375 < 10
probability histogram is / is not bell-shaped enough to use empirical rule.
6.3 The Poisson Probability Distribution
Poisson probability function, used to describe probabilities of count, X, of occurrences (successes) of an event in a specified time (or space) interval, is
P (x) = e−λt^
(λt)x x!
, x = 0, 1 ,... ,
(^2) Recall, empirical rule says 68%, 95% and 99.7% inside μ ± σ, μ ± 2 σ and μ ± 3 σ, respectively.
Empirical rule allows us to say, for example, observation is unusual if outside μ ± 2 σ.
114 Chapter 6. Discrete Probability Distributions (Lecture Notes 6)
(c) Chance at most x = 2 photons hit field in one microsecond
P (X ≤ 2) =
e−(5·1)^ (5 · 1)^0 0!
e−(5·1)^ (5 · 1)^1 1!
e−(5·1)^ (5 · 1)^2 2!
(circle one) 0. 08 / 0. 12 / 0. 18. (StatCrunch: Stat, Calculators, Poisson, Mean: 5, Prob(X ≤ 2) Compute.) (d) Chance at least x = 2 photons hit field in one microsecond
P (X ≥ 2) =
e−(5·1)^ (5 · 1)^2 2!
e−(5·1)^ (5 · 1)^3 3!
- · · · forever = 1 − P (X < 2) = 1 − P (X ≤ 1) = 1 − [P (0) + P (1)]
= 1 −
[ e−(5·1)^ (5 · 1)^0 0!
e−(5·1)^ (5 · 1)^1 1!
] ≈
(circle one) 0. 91 / 0. 93 / 0. 96. (StatCrunch: Stat, Calculators, Poisson, Mean: 5, Prob(X ≥ 2) Compute.) (e) Chance x = 2 photons hit field in two microseconds
P (2) = e−λt^
(λt)x x!
e−(5·2)^ (5 · 2)^2 2!
(circle one) 0. 002 / 0. 005 / 0. 006. (Since μ = 5 × 2 = 10, Stat, Calculators, Poisson, Mean: 10, Prob(X = 2) Compute.) (f) Chance at most x = 3 photons hit field in two microseconds
P (X ≤ 3) =
e−(5·2)^ (5 · 2)^0 0!
e−(5·2)^ (5 · 2)^1 1!
e−(5·2)^ (5 · 2)^2 2!
e−(5·2)^ (5 · 2)^3 3!
(circle one) 0. 01 / 0. 06 / 0. 08. (Since μ = 5 × 2 = 10, Stat, Calculators, Poisson, Mean: 10, Prob(X ≤ 3) Compute.) (g) Chance at least x = 21 photons hit field in four microseconds
P (X ≥ 21) = 1 − P (X < 21) = 1 − P (X ≤ 20) ≈
(circle one) 0. 44 / 0. 52 / 0. 68. (Since μ = 5 × 4 = 20, Stat, Calculators, Poisson, Mean: 20, Prob(X ≥ 21) Compute.) (h) Expected number of photon hits in one microsecond. μ = λt = 5 × 1 = (circle one) 3 / 4 / 5. (i) Standard deviation in number of photon hits in one microsecond. σ =
λt =
5 × 1 ≈ (circle one) 2. 13 / 2. 24 / 2. 45. So, expect 5 “±” 2.24 hits in one microsecond.
Section 4. The Poisson Probability Distribution (Lecture Notes 6) 115
(j) Expected number of photon hits in three microseconds. μ = λt = 5 × 3 = (circle one) 10 / 15 / 20. (k) Standard deviation in number of photon hits in three microsecond. σ =
λt =
5 × 3 ≈ (circle one) 3. 13 / 3. 24 / 3. 87. So, expect 15 “±” 3.87 hits in three microseconds. (l) Poisson process? Match columns.
Poisson process photon example (a) zero chance more than one success in small subinterval (A) assume “zero” chance two hits at same time (b) chance of success same if two intervals are of equal length (B) assume chance number hits same per microsecond (c) independence of successes in different intervals (C) assume hits independent of one another Poisson process (a) (b) (c) photon example
- Poisson: bryozoan count. During a biology study, 250 bryozoans are found attached on a submerged 10 centimeter by 10 centimeter (100 centimeters^2 ) plate, an average of λ = 250100 = 2 .5 bryozoans per one centimeter^2. Assume this is a Poisson process.
(a) Chance x = 0 bryozoans attached in one centimeter^2
P (0) = e−λt^
(λt)x x!
e−(2.^5 ·1)^ (2. 5 · 1)^0 0!
(circle one) 0. 07 / 0. 08 / 0. 09. (Stat, Calculators, Poisson, Mean: 2.5, Prob(X = 0) Compute.) (b) Chance at most x = 2 bryozoans attached in one centimeter^2
P (X ≤ 2) =
e−(2.^5 ·1)^ (2. 5 · 1)^0 0!
e−(2.^5 ·1)^ (2. 5 · 1)^1 1!
e−(2.^5 ·1)^ (2. 5 · 1)^2 2!
(circle one) 0. 54 / 0. 62 / 0. 78. (Stat, Calculators, Poisson, Mean: 2.5, Prob(X ≤ 2) Compute.) (c) Chance between x = 2 and x = 4 bryozoans in one centimeter^2 , inclusive:
P (2 ≤ X ≤ 4) = P (X ≤ 4) − P (X < 2) = P (X ≤ 4) − P (X ≤ 1) ≈
(circle one) 0. 60 / 0. 73 / 0. 76. (Stat, Calculators, Poisson, Between, Mean: 2.5, Prob(2 ≤ X ≤ 4) Compute.) (d) Chance between x = 2 and x = 4 bryozoans in three centimeters^2 , inclusive:
P (2 ≤ X ≤ 4) = P (X ≤ 4) − P (X < 2) = P (X ≤ 4) − P (X ≤ 1) ≈
(circle one) 0. 13 / 0. 23 / 0. 36. (Stat, Calculators, Poisson, Between, Mean: 7.5, Prob(2 ≤ X ≤ 4) Compute.)
