Probabilit y Assignment Uson, Adriana Co
CHAPTER 4 Exercise 4.48
The The owner owner of a restau restaura rant nt servin serving g Conti Continen nental tal-- style style entre entrees es was was interested in studying ordering patterns of patrons for the Friday to Sunday weekend time period. Records were maintained that indicated the demand for dessert during the same time period. The owner decided that two other variables were to be studied along with whether a dessert was ordered: the gender of the individual and whether a beef entrée was ordered. The results are as follows: Gender Dessert Ordered Yes No Total
Male 96 224 320
Female 40 240 280
Total 136 454 600
Beer Entrée Desert Ordered Yes No Total
Yes 71 116 187
No 65 348 413
Total 136 464 600
A waiter approaches a table to take an order. What is the probability that the first customer to order at the table: a.)
Orders a dessert?
P (D) = 0.227
b.)) b.
136+136/1200
Does Does not not ord order er a b bee eeff ent entré rée? e?
P (B’) = 0.688
413/600
c.)Orders a dessert or a beef entrée?
P (D or B) = 0.42 136+187-71/600 d.)
Is a female and does not order a dessert?
P (F and D’) = 0.40
e.)
240/600
Orders a dessert and a beef entrée?
P (D and B) = 0.118
71/600
f.) Is a female or does not order a dessert?
P (F or D’) = 0.83 280+454-240/600
g.) Suppose the 1st person that the waiter takes the dessert
order from is a female. What is the probability that she does not order a dessert?
P (D/F) = 0.857
240/280
h.) Suppose that the first person that the waiter takes the dessert order from ordered a beef entrée. What is the probability that this person orders dessert?
P (D/ B) = 0.3797 71/187
i.) Are gender and ordering dessert statistically independent?
P (M/D)= P(D)
P (M/D) = 96/600 = 0.71
136/600
P(M) = 320 = 0.53 600
No. Gender and ordering of dessert are not statistically independent because P (M/D) is not equal to P (D).
j.) Is ordering a beef entrée statistically independent of whether the person orders dessert?
P (B/D) = P (B) P (B/D) = 71/600 = 0.52 136/600 P (B) = 187 = 0.312 600 No. Ordering a beef entrée is not statistically independent with whether a person orders dessert or not because P (B/D) is not equal to P (B).
Probabilities Calculations Sample Space
Gender Male
Dessert Ordered
YES NO Totals
Simple Probabilities P(YES)
96 224 320
Female
Totals
40 240 280
136 454 600
Joint Probabilities 0.23
P(YES and Male)
0.16
P(NO)
0.76
P(YES and Female)
0.07
P(Male)
0.53
P(NO and Male)
0.37
P(Female)
0.47
P(NO and Female)
0.40
Addition Rule P(YES or Male)
0.60
P(YES or Female)
0.63
P(NO or Male)
0.92
P(NO or Female)
0.82
Probabilities Calculations Sample Space
Beef Entrée YES B
Dessert
YES D
NO B
Totals
71
65
136
NO D
116
348
464
Totals
187
413
600
Simple Probabilities
Joint Probabilities
P(YES D)
0.23
P(YES D and YES B)
0.12
P(NO D)
0.77
P(YES D and NO B)
0.11
P(YES B)
0.31
P(NO D and YES B)
0.19
P(NO B)
0.69
P(NO D and NO B)
0.58
Addition Rule P(YES D or YES B)
0.42
P(YES D or NO B)
0.81
P(NO D or YES B)
0.89
P(NO D or NO B)
0.88
CHAPTER 5 Exercise 5.47
Cinema advertising is increasing. Normally 60 to 90 seconds long, these advertisements are longer and more extravagant, and tend to have more captive audiences than television advertisements. Thus, it is not surprising that the recall rates for cinema advertisements are higher than for television advertisements. According to survey research conducted by the
ComQUEST division of BBM, the probability a viewer will remember a cinema advertisement is 0.74, whereas the probability a viewer will remember a 30second television ad is 0.37. a.) Is the 0.74 probability reported by the BBM best classified as a priori classical probability, empirical classical probability, or subjective probability?
It is classified as an empirical classical probability because the probability was obtained through a survey conducted by BBM. In the said approach, although the probability is still defined objectively as the ratio of the number of favorable outcomes to the total number of outcomes, the outcomes are based on an observed data, not on prior knowledge of a process. Hence it cannot be classified as an a priori classical probability approach. Neither can it be classified as a subjective probability because it is computed objectively by means of a survey. b.) Suppose that 10 viewers of a cinema advertisement are
randomly sampled. Consider the random variable defined by the number of viewers that recall the advertisement. What assumptions must be made in order to assume that this random variable is distributed as a binomial random variable?
The assumptions needed are: a.) the probability of each returned response is the same and b.) the result of one returned response does not affect the result of the other.
c.)Assuming that the number of viewers that recall the cinema advertisement is a binomial random variable, what are the mean and standard deviation of this distribution?
µ=NP 10(0.74)10 * 0.74 = 7.4 10 (. 74 )(. 26 )
= 1.387
d.) Based on your answer to c, if none of the viewers can recall
the ad, what can be inferred about the 0.74 probability given in the article?
P(X=0) = 10!/ 0!(10-0)!(.74) 0 (1-.74) 10-0
= (.26) 10 There is 1.4116 chance that none of the viewers will recall the cinema advertisement.
X
P(X)
P(<=X)
P(
P(>X)
P(>=X)
0
1.41E-06
1.41E-06
0
0.999999
1
1
4.02E-05
4.16E-05
1.41E-06
0.999958
0.999999
2
0.000515
0.000556
4.16E-05
0.999444
0.999958
3
0.003906
0.004462
0.000556
0.995538
0.999444
4
0.019453
0.023915
0.004462
0.976085
0.995538
5
0.066439
0.090354
0.023915
0.909646
0.976085
6
0.157581
0.247935
0.090354
0.752065
0.909646
7
0.256285
0.50422
0.247935
0.49578
0.752065
8
0.273535
0.777755
0.50422
0.222245
0.49578
9
0.173005
0.95076
0.777755
0.04924
0.222245
10
0.04924
1
0.95076
0
0.04924
Compute the probability that of the 10 viewers: e.)
Exactly zero can recall the advertisement.
P (0) is 1.4117 f.) Exactly one can recall the advertisement.
P (1) is 4.0178 g.)
Exactly two can recall the advertisement.
P (2) is 0.0005 h.)
All 10 can recall the advertisement.
P (10) is 0.0492 i.) More than half can recall the advertisement.
P (>5) is 0.9096 j.) Eight or more can recall the advertisement.
P (>=8) is 0.4957 Cinema Ad Recall Data Sample size Probability of success
10 0.74
Statistics Mean Variance Standard deviation
7.4 1.924 1.387083
Binomial Probabilities Table X 0 1 2 3 4 5 6 7 8 9 10
k.)
P(X) 1.4117E-06 4.0178E-05 0.00051459 0.00390562 0.01945299 0.06643943 0.15758069 0.25628508 0.27353504 0.17300507 0.0492399
P(<=X) 1.41E-06 4.16E-05 0.000556 0.004462 0.023915 0.090354 0.247935 0.50422 0.777755 0.95076 1
P(
P(>X) 0.999999 0.999958 0.999444 0.995538 0.976085 0.909646 0.752065 0.49578 0.222245 0.04924 0
P(>=X) 1 0.999999 0.999958 0.999444 0.995538 0.976085 0.909646 0.752065 0.49578 0.222245 0.04924
Repeat c, (e-j) for a television advertisement using the given probability of recall, 0.37.
l.) Assuming
that the number of viewers that recall the television advertisement is a binomial random variable, what are the mean and standard deviation of this distribution?
µ=NP 10(0.37)10 * 0.37 = 3.7 10 (. 37 )(. 63 )
= 1.5267
Compute the probability that of the 10 viewers:
m.)
Exactly zero can recall the advertisement.
P (0) is 0.0098 n.)
Exactly one can recall the advertisement.
P (1) is 0.0578 o.)
Exactly two can recall the advertisement.
P (2) is 0.1528 p.)
All 10 can recall the advertisement.
P (10) is 4.8085 q.)
More than half can recall the advertisement.
P (>5) is 0.1205 r.) Eight or more can recall the advertisement.
P (>=8) is 0.00714 TV Ad Recall Data Sample size Probability of success
10 0.37
Statistics Mean Variance Standard deviation
X 0 1 2 3 4 5 6 7 8 9 10
3.7 2.331 1.526761
Binomial Probabilities Table P(X) P(<=X) P(
P(>X) 0.990151 0.932306 0.779429 0.540004 0.293928 0.120503 0.035625 0.00714 0.000867 4.81E-05 0
P(>=X) 1 0.990151 0.932306 0.779429 0.540004 0.293928 0.120503 0.035625 0.00714 0.000867 4.81E-05
Exercise 5.50
To measure radio listenership, Arbitron mails surveys to randomly selected households in 283 markets in the US. The respondents are given diaries to record the radio stations they listen to. When the diaries are complete, the respondents return them to Arbitron and they receive a cash award for their participation (currently $10 in most markets). According to an article, responses to the diary survey hits a 20- year low in 2002, with only 32.6 % being returned. Suppose that in a particular market, 100 surveys are sent. a.) What assumptions do you need to make in order to use the binomial distribution to model the number of surveys returned?
The assumptions needed to model the number of surveys returned by the binomial distribution are: a.) the probability of each returned response is the same and b.) the result of one returned response does not affect the result of the other. b.) What is the expected value or mean of the binomial distribution?
µ = NP 100 * 32.6% = 32.6 c.)What is the standard deviation of the binomial distribution? 100 (. 326 )(. 674 )
= 4.6875
d.) What is the probability that 30 or fewer surveys will be returned?
P (<=30) is 0.3308 e.) What is the probability that 25 or fewer surveys will be returned?
P (<=25) is 0.0626 f.) What is the probability that more than 40 surveys will be returned?
P (>40) is 0.0478
g.) What is the probability that at least 30 but no more than 35 surveys will be returned?
P (>=30) – P (>35) = 0.4778
Binomial Distributio for Exercise 5.50 Data Sample size Probability of success
100 0.326
Statistics Mean Variance Standard deviation
X 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28
P(X) 3.55E-16 8.51E-15 1.34E-13 1.58E-12 1.46E-11 1.12E-10 7.28E-10 4.09E-09 2.02E-08 8.91E-08 3.53E-07 1 .26E-06 4.14E-06 1.24E-05 3.45E-05 8.87E-05 0.000212 0.000473 0.000987 0.001933 0.003562 0.006187 0.010149 0.015749 0.023157 0.032309 0.04283 0.05401
32.6 21.9724 4.687473
Binomial Probabilities Table P(<=X) P(X) 3.63E-16 7.34E-18 1 8.87E-15 3.63E-16 1 1.43E-13 8.87E-15 1 1.72E-12 1.43E-13 1 1.64E-11 1.72E-12 1 1.28E-10 1.64E-11 1 8.57E-10 1.28E-10 1 4.95E-09 8.57E-10 1 2.52E-08 4.95E-09 1 1.14E-07 2.52E-08 1 4.67E-07 1.14E-07 1 1.73E-06 4.67E-07 0.999998 5.87E-06 1.73E-06 0.999994 1.83E-05 5.87E-06 0.999982 5.28E-05 1.83E-05 0.999947 0.000142 5.28E-05 0.999858 0.000354 0.000142 0.999646 0.000826 0.000354 0.999174 0.001813 0.000826 0.998187 0.003747 0.001813 0.996253 0.007309 0.003747 0.992691 0.013496 0.007309 0.986504 0.023645 0.013496 0.976355 0.039393 0.023645 0.960607 0.06255 0.039393 0.93745 0.094859 0.06255 0.905141 0.137689 0.094859 0.862311 0.191699 0.137689 0.808301
P(>=X) 1 1 1 1 1 1 1 1 1 1 1 1 0.999998 0.999994 0.999982 0.999947 0.999858 0.999646 0.999174 0.998187 0.996253 0.992691 0.986504 0.976355 0.960607 0.93745 0.905141 0.862311
29 30 31 32 33 34 35 36 37 38 39 40
0.064858 0.074244 0.081087 0.084569 0.084287 0.080337 0.073274 0.063991 0.053537 0.042931 0.033011 0.024349
0.256557 0.330801 0.411888 0.496457 0.580745 0.661082 0.734356 0.798347 0.851884 0.894815 0.927825 0.952174
0.191699 0.256557 0.330801 0.411888 0.496457 0.580745 0.661082 0.734356 0.798347 0.851884 0.894815 0.927825
0.743443 0.669199 0.588112 0.503543 0.419255 0.338918 0.265644 0.201653 0.148116 0.105185 0.072175 0.047826
0.808301 0.743443 0.669199 0.588112 0.503543 0.419255 0.338918 0.265644 0.201653 0.148116 0.105185 0.072175
h.) If Arbitron increases the cash incentive so that the response rate is 40%, what are the answers to (A-f)? Use excel for (d-h). i.) What assumptions do you need to make in order to use the binomial distribution to model the number of surveys returned?
The assumptions needed to model the number of surveys returned by the binomial distribution are: a.) the probability of each returned response is the same and b.) the result of one returned response does not affect the result of the other. j.) What is the expected value or mean of the binomial distribution?
µ = NP 100 * 40% = 40 k.)
What is the distribution? 100 (. 40 )(. 60 )
standard
deviation
of
the
binomial
= 4.8990
l.) What is the probability that 30 or fewer surveys will be returned? P (<=30) is 0.0248
m.) What is the probability that 25 or fewer surveys will be returned? P(<=25) is 0.0012
n.) What is the probability that more than 40 surveys will be returned? P(>40) is 0.4567 o.) What is the probability that at least 30 but no more than 35
surveys will be returned?
P (>=30) - P(>35) = 0.1647 Data Sample size Probability of success
100 0.4
Statistics Mean Variance Standard deviation
X 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
P(X) 4.36E-21 1.44E-19 3.13E-18 5.06E-17 6.48E-16 6.84E-15 6.12E-14 4.74E-13 3.23E-12 1.96E-11 1.07E-10 5.29E-10 2.39E-09 9.89E-09 3.78E-08 1.34E-07 4 .41E-07 1.36E-06 3.9E-06 1.05E-05 2.67E-05 6.4E-05 0.000145 0.00031 0.000627 0.001207 0.002205
40 24 4.898979
Binomial Probabilities Table P(<=X) P(X) 4.42E-21 6.53E-23 1 1.48E-19 4.42E-21 1 3.28E-18 1.48E-19 1 5.39E-17 3.28E-18 1 7.02E-16 5.39E-17 1 7.54E-15 7.02E-16 1 6.87E-14 7.54E-15 1 5.43E-13 6.87E-14 1 3.78E-12 5.43E-13 1 2.34E-11 3.78E-12 1 1.3E-10 2.34E-11 1 6.59E-10 1.3E-10 1 3.05E-09 6.59E-10 1 1.29E-08 3.05E-09 1 5.07E-08 1.29E-08 1 1.85E-07 5.07E-08 1 6.26E-07 1.85E-07 0.999999 1.98E-06 6.26E-07 0.999998 5.88E-06 1.98E-06 0.999994 1.64E-05 5.88E-06 0.999984 4.32E-05 1.64E-05 0.999957 0.000107 4.32E-05 0.999893 0.000252 0.000107 0.999748 0.000562 0.000252 0.999438 0.001189 0.000562 0.998811 0.002396 0.001189 0.997604 0.0046 0.002396 0.9954
P(>=X) 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0.999999 0.999998 0.999994 0.999984 0.999957 0.999893 0.999748 0.999438 0.998811 0.997604
28 29 30 31 32 33 34 35 36 37 38 39 40
0.003832 0.006343 0.010008 0.015065 0.021656 0.02975 0.039083 0.049133 0.059141 0.068199 0.075378 0.079888 0.081219
0.008433 0.014775 0.024783 0.039848 0.061504 0.091254 0.130337 0.179469 0.238611 0.30681 0.382188 0.462075 0.543294
0.0046 0.008433 0.014775 0.024783 0.039848 0.061504 0.091254 0.130337 0.179469 0.238611 0.30681 0.382188 0.462075
0.991567 0.985225 0.975217 0.960152 0.938496 0.908746 0.869663 0.820531 0.761389 0.69319 0.617812 0.537925 0.456706
0.9954 0.991567 0.985225 0.975217 0.960152 0.938496 0.908746 0.869663 0.820531 0.761389 0.69319 0.617812 0.537925
Chapter 6 Exercise 6.75
The fill amount of bottles of soft drink has been found to be normally distributed with a mean of 2.0 liters and a standard deviation of 0.05 liter. Bottles that contain less than 95% of the listed net content (1.90 liters in this case) can make the manufacturer subject to penalty by the state office of consumer affairs, whereas bottles that have a net content above 2.10 liters may cause excess spillage upon opening. What proportion of the bottles will contain. a.)
Between 1.90 and 2.0 liters?
Z= X - µ/sd
Z= 2.0- 2.0/ 0.05 is 0 Z= 1.90- 2.0/ 0.05 is -2 From Table of Cumulative Standardized Normal
Distribution 0.5000- 0.0228 = 0.4772 or 48% Probability for a Range From X Value 1.9 To X Value 2 Z Value for 1.9 -2 Z Value for 2 0 P(X<=1.9) 0.0228 P(X<=2) 0.5000 P(1.9<=X<=2) 0.4772
b.)
Between 1.90 and 2.10 liters?
Z= X - µ/sd
Z= 2.10- 2.0/ 0.05 is 2 Z= 1.90- 2.0/ 0.05 is -2 From Table of Cumulative Standardized Normal
Distribution 0.9772- 0.0228 = 0.9545 or 95% Probability for a Range From X Value 1.9 To X Value 2.1 Z Value for 1.9 -2 Z Value for 2.1 2 P(X<=1.9) 0.0228 P(X<=2.1) 0.9772 P(1.9<=X<=2.1) 0.9545
c.)Below 1.90 liters?
Z= X - µ/sd
Z= 1.90-2.0/0.05 is -2 Z= 2%
d.)
Below 1.90 liters or above 2.0 liters?
Probability of that liters is between 1.90 and 2.0 is 0.4772, then 1-0.4772= 0.5228 or 52% e.)
Above 2.0 liters?
Z= X - µ/sd
Z= 2- 2.0/0.05 is 0; 0.5000 Complement is 1-0.5000 = 0.5000 or 50% X Value Z Value P(X>2)
2 0 0.5000
f.) Between 2.05 and 2.10 liters?
Z= X - µ/sd
Z= 2.10- 2.0/ 0.05 is 2 Z= 2.05- 2.0/ 0.05 is 1
From Table of Cumulative Standardized Normal Distribution 0.9772- 0.8413 = 0.1359 or 14% Probability for a Range From X Value 2.05 To X Value 2.1 Z Value for 2.05 1 Z Value for 2.1 2 P(X<=2.05) 0.8413 P(X<=2.1) 0.9772 P(2.05<=X<=2.1) 0.1359
g.) 99% of the bottles would be expected to contain at least how much soft drink?
X= µ+Z(sd)
X= 2 + 2.33 (0.05) X= 2.116 liters Find X and Z Given Cum. Pctage. Cumulative Percentage 99.00% Z Value 2.326348 X Value 2.116317
h.) 99% of the bottles would be expected to contain an amount
that is between distributed)?
which
two
values
(symmetrically
XL is 0.005, Z= -2.57
XU is 0.9950, Z= 2.58
XL = µ + Z(sd)
XU= µ + Z(sd)
= 2 + (-2.57) (0.05) X L= 1.87 liters
= 2 + (2.58) (0.05) X u= 2.129 liters
i.) Explain the difference in the results in g and h.
In (g), 99% of the bottles would be expected to contain at least 2.116 liters, whereas (h) explains that 99% of the bottles would be expected to contain an amount that is between 1.87 to 2.129 liters.
The difference is that in the latter, it is assumed that the data is symmetrically distributed. j.) Suppose that it an effort to reduce the number of bottles that contain less than 1.90 liters, the bottlers set the filling machine so that the mean is 2.02 liters. Under these circumstance, what would be your answer in a- h? k.)
Between 1.90 and 2.0 liters?
Z= X - µ/sd
Z= 2.0- 2.02/ 0.05 is -0.4 Z= 1.90- 2.02/ 0.05 is -2.4 From Table of Cumulative Standardized Normal
Distribution 0.3446- 0.0082 = 0.3364 or 34% Probability for a Range From X Value 1.9 To X Value 2 Z Value for 1.9 -2.4 Z Value for 2 -0.4 P(X<=1.9) 0.0082 P(X<=2) 0.3446 P(1.9<=X<=2) 0.3364
l.) Between 1.90 and 2.10 liters?
Z= X - µ/sd
Z= 2.10- 2.02/ 0.05 is 1.6 Z= 1.90- 2.02/ 0.05 is -2.4 From Table of Cumulative Standardized Normal
Distribution 0.9452- 0.0082 = 0.937 or 94% Probability for a Range From X Value 1.9 To X Value 2.1 Z Value for 1.9 -2.4 Z Value for 2.1 1.6 P(X<=1.9) 0.0082 P(X<=2.1) 0.9452 P(1.9<=X<=2.1) 0.9370
m.)
Below 1.90 liters?
Z= X - µ/sd
Z= 1.90-2.02/0.05 is -2.4 Z= 0.0082
n.)
Below 1.90 liters or above 2.0 liters?
Probability of that liters is between 1.90 and 2.0 is 0.3364, then 1-0.3364= 0.6636 or 66% o.)
Above 2.0 liters?
Z= X - µ/sd
Z= 2- 2.02/0.05 is -0.4; 0.3446 Complement is 1-0.3446 = 0.6555 or 66% Probability for X > X Value 2 Z Value -0.4 P(X>2) 0.6554
p.)
Between 2.05 and 2.10 liters?
Z= X - µ/sd
Z= 2.10- 2.02/ 0.05 is 1.6 Z= 2.05- 2.02/ 0.05 is 0.6 From Table of Cumulative Standardized Normal
Distribution 0.9452- 0.7257 = 0.2195 or 21% Probability for a Range From X Value 2.05 To X Value 2.1 Z Value for 2.05 0.6 Z Value for 2.1 1.6 P(X<=2.05) 0.7257 P(X<=2.1) 0.9452 P(2.05<=X<=2.1) 0.2195
q.) 99% of the bottles would be expected to contain at least how much soft drink?
X= µ+Z(sd)
X= 2.02 + 2.33 (0.05) X= 2.136 liters Find X and Z Given Cum. Pctage. Cumulative Percentage 99.00% Z Value 2.326348 X Value 2.136317
r.) 99% of the bottles would be expected to contain an amount that is between which two values (symmetrically distributed)?
XL is 0.005, Z= -2.57
XU is 0.9950, Z= 2.58
XL = µ + Z(sd)
XU= µ + Z(sd)
= 2.02 + (-2.57) (0.05) X L= 1.8915 liters
= 2.02 + (2.58) (0.05) X u= 2.149 liters
s.)Explain the difference in the results in g and h.
In (g), 99% of the bottles would be expected to contain at least 2.136 liters, whereas (h) explains that 99% of the bottles would be expected to contain an amount that is between 1.8915 to 2.149 liters. The difference is that in the latter, it is assumed that the data is symmetrically distributed. If a random sample of 25 bottles is selected, what is the probability that the sample mean will be t.) Between 1.99 and 2.0 liters?
Z= X–μ σ/√n
= 2.0- 2.0 .05/√25
=0 Z= X–μ σ/√n
= 1.99- 2.0 .05/√25
= -0.001
0.01 = -.1 From Table of Cumulative Standardized Normal Distribution 0.5000- 0.4602 = 0.398 or 39.8%
u.)
Between 1.99 and 2.01 liters?
Z= X–μ σ/√n
= 2.01- 2.0 .05/√25
=1 Z= X–μ σ/√n
= 1.99- 2.0 .05/√25
= -0.001 0.01 = -.1 From Table of Cumulative Standardized Normal Distribution 0.8413- 0.4602 = 0.3811 or 38.11% v.)Below 1.98 liters?
Z = X – μ= 1.98- 2.0 σ/√n .05/√25 = -0.02 0.01 = -2; or 0.0228 or 2.28% of the sample size 25 would be expected to have means below 1.98L
w.)
Below 1.98 liters or above 2.02 liters?
Z= X–μ σ/√n
= 1.98- 2.0 .05/√25
= -0.02 0.01
= -2 or 0.0228 or 2.28% Z= X–μ σ/√n
= 2.02- 2.0 .05/√25
= 0.02 0.01 = 2; or 0.9772 or 97.72% Complement is 1- 0.9772 = 0.0228 0.0228 + 0.9772 = 1 x.)Above 2.01 liters?
Z= X–μ σ/√n
= 2.01- 2.0 .05/√25
= 0.01 0.01 = 1; or 0.8413 or 84.13% Complement is 1- 0.8413 = 0.1587 or 15.87% is above 2.01 liters y.)Between 2.01 and 2.03 liters?
Z= X–μ σ/√n
= 2.01- 2.0 .05/√25
=1 Z= X–μ σ/√n
= 2.03- 2.0 .05/√25
= -0.03 0.01 = -3 From Table of Cumulative Standardized Normal Distribution 0.8413- 0.00135 = 0.83995 or 83.99%
z.) 99% of the sample means would be expected to contain at
least how much soft drink?
2.023263 Find X and Z Given Cum. Pctage. Cumulative Percentage
99.00%
Z Value
2.326348
X Value
2.023263
99% of the sample means would be expected to contain an amount that is between which two values (symmetrically distributed around the mean?)
aa.)
XL is 0.05, Z= -2.58 XL = µ + Z(.40/
25
XU is 0.995, Z= +2.58 )
= 2.0 + (-2.58) (0.08) X L= 1.7936 liters
XU= µ + Z(.40/
25
)
= 4.70 + (+2.58) (0.08) X u= 2.2064 ounces
bb.) Explain the difference in results in q and r.
In (q), 99% of the bottles would be expected to contain at least 2.0232 liters, whereas (r) explains that 99% of the bottles would be expected to contain an amount that is between 1.7936 to 2.2064 liters. The difference is that in the latter, it is assumed that the data is symmetrically distributed.
Exercise 6.76
An orange producer buys all his oranges from a large orange grove. The amount of juice squeezed from each of these oranges is approximately normally distributed with a mean of 4.70 ounces and a standard deviation of 0.40 ounce.
a.) What is the probability that a randomly selected orange will contain between 4.70 and 5.00 ounces?
Z= X - µ/sd
Z= 5.00- 4.70/ 0.40 is 0.75 Z= 4.70- 4.70/ 0.40 is 0 From Table of Cumulative Standardized Normal
Distribution 0.7734- 0.5000 = 0.2734 Probability for a Range From X Value 4.7 To X Value 5 Z Value for 4.7 0 Z Value for 5 0.75 P(X<=4.7) 0.5000 P(X<=5) 0.7734 P(4.7<=X<=5) 0.2734
b.) What is the probability that a randomly selected orange will contain between 5.00 and 5.50 ounces?
Z= X - µ/sd
Z= 5.50- 4.70/ 0.40 is 2 Z= 5.00- 4.70/ 0.40 is 0.75 From Table of Cumulative Standardized Normal
Distribution 0.9772- 0.7734 = 0.2038 Probability for a Range From X Value 5 To X Value 5.5 Z Value for 5 0.75 Z Value for 5.5 2 P(X<=5) 0.7734 P(X<=5.5) 0.9772 P(5<=X<=5.5) 0.2039
c.)77% of the oranges will contain at least how many ounces of juice?
X= µ+Z(sd)
X= 4.70 + 0.74 (0.40)
X = 4.996 ounces Find X and Z Given Cum. Pctage. Cumulative Percentage 77.00% Z Value 0.738846849 X Value 4.99553874
d.) Between what two values (in ounces) symmetrically distributed around the population mean will 80 % of the oranges fall? XL is 0.10, Z= - 1.26 XL = µ + Z(sd) = 4.70 + (-1.26) (0.40) X L= 4.19 ounces
XU is 0.90, Z= +1.26 XU= µ + Z(sd) = 4.70 + (+1.26) (0.40) X u= 5.212 ounces
Suppose that a sample of 25 oranges is selected: e.) What is the probability that the sample mean will be at least
4.60 ounces?
_ Z = X – μ = 4.70- 4.60 σ/√n .4/√25 = 0.1 0.08 = 1.25; or 0.8944 89.44% of all the possible samples of size 25 have a sample mean of at least 4.60 ounces.
f.) Between what two values symmetrically distributed around the population mean will 70% of the sample mean fall? XL is 0.15, Z= -1.04
XU is 0.85, Z= +1.04
XL = µ + Z(.40/
25
)
XU= µ + Z(.40/
= 4.70 + (-1.04) (0.08) X L= 4.6168 ounces
g.)
25
)
= 4.70 + (+1.04) (0.08) X u= 4.7832 ounces
77% of the sample means will be above what value?
X= µ+Z(sd)
X= 4.70 + -0.74 (0.08) X = 4.6408 ounces