Download (Random Variables): Review of 5.1

Review 5.1 - 5.2 (regular Stats) - Random Variables Intro (Prob Distn) & Binomial Prob
Ch. 5 (Random Variables): Review of 5.1 - 5.2
(1)
a)
Name:
Find the missing value to make this a valid probability distribution
x
0
1
2
3
4
p(x)
0.20
0.15
0.35
0.20
0.10
b) For the above probability distribution, calculate mx = E(X):
E(X) = S xipi = 0.00
(2)
0.15
0.70
0.60
0.40
1.85
For the given values of the random variable X, and corresponding frequencies,
complete the calculation of the standard deviation, sx
Given:
X
0
1
2
3
Freq
5
20
40
15
80
x
p(x)
0
1
2
3
0.063
0.250
0.500
0.188
1.000
p(x)∙(x - m)2
0.205
0.165
0.018
0.264
x∙p(x) (x - E(x))2
0.00
0.25
1.00
0.56
3.29
0.66
0.04
1.41
mx = E(X) = 1.81
s2x = 0.652 = S pi∙(xi - m)2
S xipi
sx = 0.808
(3)
What are the 4 necessary conditions for using the Binomial Probability formula to calculate
the probability of obtaining exactly k successes out of n trials
• the number of trials, n, is fixed
• each trial is independent
• each trial is either a success (1) or a failure (0)
• P(success) on each trial is the same (p)
(4) The binomial probability of obtaining exactly k "successes" in n trials is given by:
k
n-k
nCk ∙ p (1-p)
a)
what does p represent?
p = probability of success on each trial
b) what does (1-p) represent ?
q = 1-p = probability of failure on each trial
c) Write the binomial coefficient, nCk , using its equivalent factorial notation
n!
nC k =
k!∙(n-k)!
Example:
40C37
=
40!
37! ∙ 3!
=
40 ∙ 39 ∙ 38
3∙2∙1
On TI-84 Calculator: 40 [math] [PRB] nCr 37
gives
=
C =
40 37
9880
9880
(5) For a given situation where the binomial probability distribution applies:
The number of trials is 40
n = 40
The probability of a "success" is 3/8 and the probability of a non-success is 5/8.
p = 0.375
Write a binomial probability expression involving the form, nCk ∙ pk(1-p)n-k, for:
a)
The probability of obtaining exactly 3 successes
b)
The probability of obtaining 7 or 8 successes
C ∙ (.375)7(.625)33
+
40 7
c)
The probability of obtaining 0 successes
C ∙ (.375)3(.625)37
40 3
C ∙ (.375)8(.625)32
0
40
40C0 ∙ (.375) (.625)
40 8
d)
The probability that you don't have exactly 20 successes
20
20
1 −
40C20 ∙ (.375) (.625)
(6) Use your calculator to find each of the above probabilities in #5
On TI84 / TI84+ Calculator: [2nd] [Distr] BinPDF [enter] n , p , k
a)
BinPDF(40, 0.375, 3) =
b) BinPDF(40, 0.375, 7) +
0.00357
c)
d)
BinPDF(40, 0.375, 0) =
0.0000146
BinPDF(40, 0.375, 8) =
+
0.00884
=
0.01241
0.0000000068
1 − BinPDF(40, 0.375, 20) =
<< 0.000 000 006 842
1 −
0.0345
=
0.9655
Chapter Review (5.1 - 5.2) continued
(7) Sam is a representative who sells large appliances such as refrigerators, stoves, etc.
Let x = number of appliances Sam sells on a given day. Let f = frequency with which
he sells x appliances. For a random sample of 240 days, Sam had the following sales
record.
x
0
1
2
3
4
5
6
7
f
9
72
63
41
28
14
8
5
240
Assume that the sales record is representative of the population of all sales days.
a)
Use the relative frequency to find P(x) for x = 0 to 7
x
p(x)
1
0.300
2
0.263
3
0.171
4
0.117
5
0.058
6
0.033
7
0.021
5
6
7
Use a histogram to graph the probability distribution of part (a).
0.400
Relative
Frequency = p(x)
b)
0
0.038
0.300
0.200
0.100
0.000
0
1
2
3
4
x (# of appliances / day)
100%
c)
Compute the probability that x is between 2 and 5 inclusive (i.e., also including 2 and 5)
Mutually Exclusive: P(2 or 3 or 4 or 5) = P(2) + P(3) + P(4) + P(5)
Add:
0.263 0.171 0.117 0.058
=
d)
Compute the probability that x is less than 3
P(0) + P(1) + P(2)
Add:
e)
0.038
0.300
0.263
=
0.600
Compute the expected value of the x distribution.
x
p(x)
f)
0.608
0
0.038
1
0.300
2
0.263
3
0.171
4
0.117
5
0.058
6
0.033
7
0.021
x ∙ p(x) = 0.000
0.300
0.525
0.513
0.467
0.292
0.200
0.146
Compute the standard deviation of the x distribution.
s2x = S pi∙(xi - m)2 =
Can use approach w/ #2 above
p(x)∙(x - m)2 =
0.224
0.624
0.051
2.44
= S xipi
0.053
0.283
0.382
0.433 2.472
0.422
sx = sqrt( s2x ) = 1.572
Or, On Calculator:
[Stat] [ Edit ] enter x-values in L1, enter p(x) values in L2
[Stat] [Calc] 1-Var Stats L1, L2
x-bar = 2.442
sx =
1.572
(8) The director of a health club conducted a survey and found that 23% of members used only the
pool for workouts. Based on this information, what is the probability that, for a random sample
of 10 members, 4 used only the pool for workouts?
n = 10 ; p = 0.230 ;
10C4 ∙ (.375)4(.625)6
k=4
BinPDF(10, 0.230, 4) =
0.1225
(9) Of those mountain climbers who attempt Mt. McKinley, only 65% reach the summit. In a random
sample of 16 mountain climbers who will attempt Mt. McKinley, what is the probability that:
a)
All 16 reach the summit
16C16 ∙ (.65)16(.35)0
Calculator: [2nd] [Distr] BinPDF [enter] n, p, k
BinPDF(16, 0.650, 16) =
b) At least 10 reach the summit
0 1 2 3 4 5
6
7
8
9
10
11
0.0010153
12
13
14
15
16
Recall, BinCDF (Cumulative Distribution Function) gives P( < k successes)
always "less than or equal to"
CDF for <
On Calculator: [2nd] [Distr] BinCDF [enter] n, p, k
P(at least 10) = P( k > 10) = 1 − P( k < 9)
= 1 − BinCDF(16, 0.650, 9)
= 1 − 0.3119
=
0.6881
c)
No more than 12 reach the summit
0 1 2 3 4 5 6
7
8
9
10
11
12
13
14
15
16
13
14
15
16
P(no more than 12) = P( k < 12) =
=
BinCDF(16, 0.650, 12)
=
0.8661
d)
From 9 to 12 reach the summit (including 9 & 12)
0
1
2
3
4
5
6
7
8
9
10
11
12
subtract CDF(… 8)
Can add Individual probabilities
=BinPDF(16, .65, 9) + BinPDF(16, .65, 10) + BinPDF(16, .65, 11) + BinPDF(16, .65, 12)
0.1524
+
0.1982
+
0.2008
Or can use Cumulative probabilities
P( 9 < k < 12) = P( k < 12) − P (k < 8)
=
BinCDF(16, 0.650, 12)
−
=
0.8661
−
+
0.1553
=
0.7067
BinCDF(16, 0.650, 8)
0.1594
=
0.7067