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STT 231
Problem set #4 (Due Mon, April 19) (Answers)
4/9/99
Please submit the problems in the order assigned.
1. R/L, Ch 10, #11, p236.
(a) Yes. n=10, p depends on the player. (b) Yes. n=10, p=.7. (c) No. It’s a waiting time rv. (d) No. Not defined on a specified set
of Bernoulli trials – no n, no p. (e) No. Without replacement leads to dependent trials. (f) Yes. n=10, p depends on die bias.
(g) Yes. n=10, p=.25. (h) No. There are no Bernoulli trials defined. (i) Yes – IF: must assume P(sale) =p for each contact.
2. R/L, Ch 10, #14, p236. Start by finding the ‘expected’ for each i. These come from the BIN(n,p) model with n=4, p=1/6:
4 1 i 5 4 i
i
0
1
2
3
4
P[X=i]=
( ) ( ) , i=1,2,3,4.
.482 .386 .116 .015 .001
i 6 6
P(i sixes)
F
IJ
G
HK
The corresponding ‘expected’ values are obtained by multiplying the probabilities by 1200:
i
0
1
2
3
4
expected 578.4 463.2 139.2
18
1.2 (Combine last 2 classes because 1.2 is too small)
observed 551
477
145
24
3
( 551  578.4)2 ( 477  463.2)2 (145  139.2)2 ( 27  19.2)2
= 5.12. Look at Chi-sq. , 3 df. This is below the 90th



578.4
463.2
139.2
19.2
percentile. The model is acceptable.
2 =
3. The batteries produced by a certain process have a defective rate of 10%.
(a) Let X be the number of defective batteries among 100 produced. Obtain (i) P[X=15], exactly; (ii) P[ 8  X  12] , using the normal
approximation using the continuity correction; (iii) the exact value of the probability in (ii) – using Minitab.
(i) From Minitab: MTB > pdf 15;
Result: Binomial with n = 100 and p = 0.100000
SUBC> binomial 100 .1.
x
P( X = x)
15.00
0.0327
(ii)Normal Approximation (with correction): P[7.5  X  12.5]  P[ -.83  Z  .83] =.5934.
(iii) Exact value – from MTB: x
P( X <= x)
12.00
0.8018
7.00
0.2061
P = difference = 0.5957. Good approximation.
(b) As batteries are produced, they are packed in boxes that hold 8 batteries. Let T8 be the number of batteries that will be produced in
order to fill one box. Use the binomial table to determine P[ T8  11].
P[ T8  11| p=.9] = P[ X10  7| p=.9] = .0702 (from table p 380).
(c) Boxes are shipped in crates that hold 2 dozen boxes. What are the expected value and standard deviation of the total number of
batteries that must be produced, in order to fill a crate.
To fill 2 dozen boxes requires 24x8 = 192 batteries. The rv T192 has mean 192/.9 = 213.3 and variance 192(.1)/.81 = 23.7.The sd
is Sqrt(23.7) = 4.87. See hand-out on waiting times for details.
4. The time (hrs) it takes to perform a certain task is a continuous rv X with the following density:
3 2
f(x) =
x , 1  x  2, and 0 elsewhere
7
(a ) Sketch this density. (b) Determine F, the cdf of X and sketch it also. The cdf for this model:
t3
t3  1
x 2 dx =
F(t)=
, 1 t  2 {also: F(t) =0 if t<1; F(t)=1 if t>2}
1 7
7
(b)
(a)
z
12/7
F(t)
f(x)
3/7
0
1
2
(c) Determine each of the following properties of X:
0
1
2
(i) the median ; (ii) the mean; (iii) the standard deviation, (iv) the 20 th percentile; (v) P[ 1.5 < X < 1.9].
2 3
3 x 4 2 3 16  1
m3  1
(i) Median: F(m)=. 5:
=. 5  m 3 =4.5 and m =1.65. (ii) E(X) =
=
=45/28=1.61.
x x 2 dx  
1
7
7 4 1 7 4
7
5
2
3
3 x 2 3 32  1
(iii) E( X 2 ) =
=
=93/35=2.66. Var(X) = 2.66- (1.61)2 =.068. SD(X) = .068 =.261.
x 2 x 2 dx =
1
7
7 5 1 7 5
t3  1
(iv) 20th %ile:
= .2  t 3 =2.4 and t = 20th percentile = 1.34.
7
(1.9)3  1 (1.5)3  1
(v) P[1.5 < X < 1.9] = F(1.9) – F(1.5) =

 .498.
7
7
z
z
]
]
5. A certain bus arrives at your stop at a time uniformly distributed between 10 and 10:30. You arrive at 10 o’clock sharp.
The waiting time, T, has a Unif[0, 30] distribution. CDF: F(t) = t/30, 0  t  30.
(a) What is the probability that you will have to wait more than 5 minutes for the bus?
P[T > 5 ] = 1-P[T  5] = 1-F(5) = 25/30 = .833
(b) What is the probability that the time you spend waiting is between 5 and 20 minutes? P[5 < T < 20]=(20-5)/30 = .5
(c) If at 10:15, the bus hasn’t yet arrived, what is the conditional probability that you will have to wait at least 10 more minutes?
5 30
P[ T > 15 + 10 | T > 15] =
=1/3
15 30
(d) In the course of a month, you face this situation 24 times. Approximate the probability that the total time you spend waiting for
the bus during the month is more than 6.5 hours. Use the normal approximation: E(T) = 15, SD(T) = 30/ 12 =8.66.
Mean of total time = 24x15 = 360; SD of total time = 24 SD(T)= 42.43. Note: 6.5 hrs = 390 min.
390  360
P[ total time > 390]  P[ Z >
= .71 ] = .2389.
42.43
6. (a) R/L, Ch 11, #9, p259. (b) Use Minitab to obtain the deciles of the filling distribution given in (a). Deciles are the following
percentiles: 10th, 20th, 30th, ….,90th - there are 9 of them. Assume X ~ N(12.1, .1).
12  12.1
11.9  12.1
(a) (i) P [X < 12 ] = P[ Z <
] = P[ Z < -1]= .1587. (ii) P X > 11.9] = P[ Z >
] = P[Z > -2] = .9772.
.1
.1
(b)
MTB > invcdf c1 c2;
Data Display
SUBC> normal 12.1 .1.
10ths
deciles 10ths deciles
MTB > print c1 c2
0.1
11.97
0.5
12.10
0.2
12.02
0.6
12.13
0.3
12.05
0.7
12.15
0.4
12.07
0.8
12.18
0.9
12.23
7. This is a modified version of R/L, Ch 11, # 31: The only change is in the data: 75 59 65 70 59 75 77 79 61 72 67 72.
S
S
The form of the interval: [ X  t
] where X is the sample mean, S is the sample estimator of the population
,Xt
n
n
standard deviation, and t is obtained from the t-table for the given sample size, n=12. The value of S is obtained from the
formula: S2 
X
 ( X  X)
i
n1
2
. The numerator is most easily calculated as:
 ( X  X)
 ( X  X)   X
2
i
2
i
 nX 2 . For this sample we
538.25
=6.995. The value of t
11
6.995
6.995
obtained from the table on page 257 is t = 2.201. The 95% CI: [69.25-(2.201)(
), 69.25+(2.201)(
)] = [ 64.81,73.69].
12
12
The interval contains   68 . Therefore the hypothesis is accepted.
have:
2
i
= 58,085 and X =69.25. Hence,
i
2
=58085 - 12( 69.25 )2 =538.25 and S =
8. Specifications call for a certain vacuum tube to have lifetime, T, with an exponential distribution whose mean is 50 hours.
(a) Determine the 25th, 50th, and 75th percentiles of this distribution. Recall: F(t) = 1- e

t
50
, t0.
From the hand-out on Exp(  ): x p =  ln(1/1-p). Here,  =50.
Percentiles: 25th
ln(1/(1-.25))=.288; ln(1/.5)=.693; ln(1/(1-.75))=1.386.
14.400
50th
75th
34.650
69.300
(b) Assume this model. Evaluate each of the following probabilities: P[T<15], P[15T<40], P[40T<60], P[60T<85], P[T 85]
P[T < 15] = 1- e

15
50

40
=.259 P[T < 40] = 1- e 50 =.551
P[T < 60]=.699
P[T < 85]=.817.
P[15<T<40] =.551-.259=.292; P[40<T<60] = .699-.551= .148; P[60<T<85] = .118; P[T85] = .183.
(c) Among 100 of these tubes tested, 20 had lifetimes less than 15 hours, 35 had lifetimes between 15 and 40 hours, 20 had lifetimes
between 40 and 60 hours, 15 had lifetimes between 60 and 85 hours and 10 had lifetimes greater than 85 hours. Are these results
consistent with the stated requirement that T~Exp(50)?
Intervals: < 15 15-40
40-60 60-85 >85
Observed :
20
35
20
15
10
Expected : 25.9 29.2
14.8
11.8 18.3

( 20  25.9)2 (35  29.2)2
(10  18.3)2
= 8.955. From Chi-sq. table with 4 df, this is near the 5% upper tail

 ...
25.9
29.2
18.3
(the 95th percentile). The fit is questionable, but the model isn’t totally discredited.
2
=
9. Minitab:
(a) Choose an integer (n ) at random from 25 to 75: Use the Calc  random data  integer menu or the commands:
random 1 c1;
integer 25 75.
(b) Pick a probability (p) at random in [.1, .9]: Use the same menu with ‘uniform’ instead of ‘integer’ or
random 1 c2;
uniform .1 .9.
(c) Let k1= c1(1) – that’s your n; let k2=c2(1) – that’s your p.
(d) Obtain 25 samples, each of size 40 from a binomial with your n and p.
(e) Obtain a summary (use the command ‘describe’) for 2 of your samples (it doesn’t matter which).
(f) For every one of the 25 samples use the appropriate Minitab commands (or menu) to determine an 80% (t) confidence interval
for  (pretend that neither  nor  is known).
(g) Edit your output for submission. Turn in in this order (i) Your n and p; (ii) part (e) above; (iii) the 25 confidence intervals;
(iv) Answer the following questions: What proportion of the confidence intervals missed their target? Was this about as expected?
Don’t clutter your paper with anything else.
Data Display
n
42.0000
p
0.417582
MTB > random 40 c1-c25;
SUBC> binomial k1 k2.
MTB > desc c1 c2
Descriptive Statistics
(i)
Variable
C1
C2
N
40
40
Mean
18.400
17.900
Median
18.500
18.000
Tr Mean
18.444
17.917
Variable
C1
Min
13.000
Max
23.000
Q1
16.000
Q3
20.750
C2
11.000
25.000
15.000
21.000
MTB > let k3=k1*k2
MTB > name k3 'mu'
MTB > print k3
Data Display
StDev
2.898
3.848
SE Mean
0.458
0.608
(ii)
mu
17.5385
(This
MTB > tint 80 c1-c25
is the target of the CI’s)
Confidence Intervals
Variable
C1
C2
C3
C4
C5
C6
C7
C8
C9
C10
C11
C12
C13
C14
C15
C16
C17
C18
C19
C20
C21
C22
C23
C24
C25
N
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
(iii)
Mean
18.400
17.900
17.450
17.350
17.475
16.625
16.525
17.650
16.925
17.350
18.275
17.500
17.600
17.800
18.025
17.725
17.275
17.950
17.775
16.950
17.225
16.050
17.225
17.300
17.100
StDev
2.898
3.848
3.748
2.905
3.320
3.232
3.929
2.905
3.354
3.150
2.978
2.801
3.327
3.023
3.662
3.121
3.013
3.250
3.779
2.855
3.198
2.917
2.304
3.695
3.327
SE Mean
0.458
0.608
0.593
0.459
0.525
0.511
0.621
0.459
0.530
0.498
0.471
0.443
0.526
0.478
0.579
0.494
0.476
0.514
0.598
0.451
0.506
0.461
0.364
0.584
0.526
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
80.0 %
17.803,
17.107,
16.677,
16.751,
16.791,
15.959,
15.715,
17.051,
16.234,
16.701,
17.661,
16.923,
16.914,
17.177,
17.270,
17.082,
16.654,
17.280,
16.996,
16.362,
16.566,
15.449,
16.750,
16.538,
16.414,
CI
18.997)
18.693)
18.223)
17.949)
18.159)
17.291)
17.335)
18.249)
17.616)
17.999)
18.889)
18.077)
18.286)
18.423)
18.780)
18.368)
17.896)
18.620)
18.554)
17.538)
17.884)
16.651)
17.700)
18.062)
17.786)
missed mu=17.5385
missed
missed
missed
missed (?)
missed
MTB >(iv)Out of 25 intervals, 3 missed on the high end(U<mu)
and 2 missed on the low end (L>mu); one was borderline. The total of
5 misses out of 25 intervals is exactly as predicted for 80% CI's.
The commands used:
MTB >
SUBC>
MTB >
MTB >
SUBC>
MTB >
MTB >
MTB >
MTB >
MTB >
MTB >
MTB >
MTB >
random 1 c1;
integer 25 75.
Let k1=c1(1)
random 1 c1;
uniform .1 .9.
let k2=c1(1)
name k1 'n'
name k2 'p'
print k1 k2
let k3=k1*k2
name k3 'mu'
print k3
tint 80 c1-c25
Random integer
Random p
Display n,p
Display mu
Obtain 25 CI’s based on the t dist.