4/9/99 STT 231
Problem set #4 (Due Mon, April 19) (Answers)
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:
P[X=i]=, i=1,2,3,4.
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
= = 5.12. Look at Chi-sq. , 3 df. This is below the 90th percentile. The model is acceptable.
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 be the number of batteries that will be produced in order to fill one box. Use the binomial table to determine P[ 11].
P[ 11| p=.9] = P[ 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 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:
f(x) = , 1 x 2, and 0 elsewhere
(a ) Sketch this density. (b) Determine F, the cdf of X and sketch it also. The cdf for this model:
F(t)==, 1 t 2 {also: F(t) =0 if t<1; F(t)=1 if t>2}
(b) (a)
12/7
F(t) f(x)
3/7
0 1 2 0 1 2
(c) Determine each of the following properties of X:
(i) the median ; (ii) the mean; (iii) the standard deviation, (iv) the 20th percentile; (v) P[ 1.5 < X < 1.9].
(i) Median: F(m)=. 5: =. 5 =4.5 and m =1.65. (ii) E(X) = ==45/28=1.61.
(iii) E() = == =93/35=2.66. Var(X) = 2.66-=.068. SD(X) = =.261.
(iv) 20th %ile: = .2 =2.4 and t = 20th percentile = 1.34.
(v) P[1.5 < X < 1.9] = F(1.9) – F(1.5) = .498.
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?
P[ T > 15 + 10 | T > 15] = =1/3
(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/=8.66.
Mean of total time = 24x15 = 360; SD of total time = SD(T)= 42.43. Note: 6.5 hrs = 390 min.
P[ total time > 390] P[ Z > = .71 ] = .2389.
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).
(a) (i) P [X < 12 ] = P[ Z < ] = P[ Z < -1]= .1587. (ii) P X > 11.9] = P[ Z > ] = P[Z > -2] = .9772.
(b) MTB > invcdf c1 c2;
SUBC> normal 12.1 .1.
MTB > print c1 c2
Data Display
10ths deciles 10ths deciles
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.
The form of the interval: [] where is the sample mean, S is the sample estimator of the population 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: . The numerator is most easily calculated as: . For this sample we have: = 58,085 and =69.25. Hence, =58085 - =538.25 and S = =6.995. The value of t 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].
The interval contains . Therefore the hypothesis is accepted.
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-, t0.
From the hand-out on Exp(): = ln(1/1-p). Here, =50. Percentiles: 25th 50th 75th
ln(1/(1-.25))=.288; ln(1/.5)=.693; ln(1/(1-.75))=1.386. 14.400 34.650 69.300
(b) Assume this model. Evaluate each of the following probabilities: P[T<15], P[15T<40], P[40T<60], P[60T<85], P[T 85]
P[T < 15] = 1-=.259 P[T < 40] = 1-=.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[T85] = .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
== 8.955. From Chi-sq. table with 4 df, this is near the 5% upper tail
(the 95th percentile). The fit is questionable, but the model isn’t totally discredited.
9. Minitab:
(a) Choose an integer (n ) at random from 25 to 75: Use the Calc random datainteger 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 (i)
MTB > random 40 c1-c25;
SUBC> binomial k1 k2.
MTB > desc c1 c2
Descriptive Statistics
Variable N Mean Median Tr Mean StDev SE Mean
C1 40 18.400 18.500 18.444 2.898 0.458
C2 40 17.900 18.000 17.917 3.848 0.608
Variable Min Max Q1 Q3
C1 13.000 23.000 16.000 20.750
C2 11.000 25.000 15.000 21.000 (ii)
MTB > let k3=k1*k2
MTB > name k3 'mu'
MTB > print k3
Data Display
mu 17.5385 (This is the target of the CI’s)
MTB > tint 80 c1-c25
Confidence Intervals
(iii)
Variable N Mean StDev SE Mean 80.0 % CI
C1 40 18.400 2.898 0.458 ( 17.803, 18.997) missed mu=17.5385
C2 40 17.900 3.848 0.608 ( 17.107, 18.693)
C3 40 17.450 3.748 0.593 ( 16.677, 18.223)
C4 40 17.350 2.905 0.459 ( 16.751, 17.949)
C5 40 17.475 3.320 0.525 ( 16.791, 18.159)
C6 40 16.625 3.232 0.511 ( 15.959, 17.291) missed
C7 40 16.525 3.929 0.621 ( 15.715, 17.335) missed
C8 40 17.650 2.905 0.459 ( 17.051, 18.249)
C9 40 16.925 3.354 0.530 ( 16.234, 17.616)
C10 40 17.350 3.150 0.498 ( 16.701, 17.999)
C11 40 18.275 2.978 0.471 ( 17.661, 18.889) missed
C12 40 17.500 2.801 0.443 ( 16.923, 18.077)
C13 40 17.600 3.327 0.526 ( 16.914, 18.286)
C14 40 17.800 3.023 0.478 ( 17.177, 18.423)
C15 40 18.025 3.662 0.579 ( 17.270, 18.780)
C16 40 17.725 3.121 0.494 ( 17.082, 18.368)
C17 40 17.275 3.013 0.476 ( 16.654, 17.896)
C18 40 17.950 3.250 0.514 ( 17.280, 18.620)
C19 40 17.775 3.779 0.598 ( 16.996, 18.554)
C20 40 16.950 2.855 0.451 ( 16.362, 17.538) missed (?)
C21 40 17.225 3.198 0.506 ( 16.566, 17.884)
C22 40 16.050 2.917 0.461 ( 15.449, 16.651) missed
C23 40 17.225 2.304 0.364 ( 16.750, 17.700)
C24 40 17.300 3.695 0.584 ( 16.538, 18.062)
C25 40 17.100 3.327 0.526 ( 16.414, 17.786)
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 > random 1 c1;
SUBC> integer 25 75. Random integer
MTB > Let k1=c1(1)
MTB > random 1 c1;
SUBC> uniform .1 .9. Random p
MTB > let k2=c1(1)
MTB > name k1 'n'
MTB > name k2 'p'
MTB > print k1 k2 Display n,p
MTB > let k3=k1*k2
MTB > name k3 'mu'
MTB > print k3 Display mu
MTB > tint 80 c1-c25 Obtain 25 CI’s based on the t dist.