Based on data from a car bumper sticker study, when a car is randomly selected,
ID: 3265413 • Letter: B
Question
Based on data from a car bumper sticker study, when a car is randomly selected, the number of bumper stickers and the corresponding probabilities are as shown below Complete parts (a) through (d) Click the icon to view the data table a. Does the given information describe a probability distribution? O No O Yes b. Assuming that a probability distribution is described, find its mean and standard deviation The mean is (Round to the nearest tenth as needed) The standard deviation is (Round to the nearest tenth as needed) c. Use the range rule of thumb to identify the range of vakues for usual numbers of bumper stickers The maximum usual value is (Round to the nearest tenth as needed ) The minimum usual value is (Round to the nearest tenth as needed) d. Is it unusual for a car to have more than one bumper sticker? Explain A. No, because the probability of more than 1 bumper sticker is 0.123, which is greater than 0 05. O B. Yes, because the probabilities for random variable x from 2 to 9 are all less than 0 05 O C. No, because the probability of having 1 bumper sticker is 0 081, which is greater than 0 05 O D. Not enough information is givenExplanation / Answer
Part (a)
Answer: YES
Because, the probability values are all non-negative decimals and sum of all probabilities is 1, two necessary conditions for a probability distribution.
Part (b)
If pi be the probability X = xi, i = 1, 2, ….., n. Then, Mean = E(X) = [1,n](xipi);
E(X2) = [1,n](xi2pi); and variance = V(X) = E(X2) – { E(X)}2
SD(X) = sqrt{V(X)}
Computatios
x
0
1
2
3
4
5
6
7
8
9
p(x)
0.796
0.08
0.05
0.02
0.01
0.01
0.01
0.01
0.01
0.00
x.p(x)
0
0.081
0.092
0.072
0.056
0.07
0.054
0.042
0.048
0.036
x^2.p(x)
0
0.081
0.184
0.216
0.224
0.35
0.324
0.294
0.384
0.324
E(X) =
0.551
E(X^2) =
1.673
V(X) =
1.3694
SD(X) =
1.17021
ANSWER: Mean = 0.6, SD = 1.2
Part (c)
Maximum value = Mean + 3SD = 4.2 and Minimum value = Mean - 3SD = -2.6 = 0 [since negative values for number of bumper stickesr]
ANSWER: max = 4.2 and min = 0
Part (d)
Probability of more than 1 bumper sticker = P(X > 1), where X = number of bumper stickers in a car.
= 0.123 which is more than the normal acceptable value of 0.5. So, it is NOT unusual for a car to have more than ANSWER First Option
If pi be the probability X = xi, i = 1, 2, ….., n. Then, Mean = E(X) = [1,n](xipi);
E(X2) = [1,n](xi2pi); and variance = V(X) = E(X2) – { E(X)}2
SD(X) = sqrt{V(X)}
Computatios
x
0
1
2
3
4
5
6
7
8
9
p(x)
0.796
0.08
0.05
0.02
0.01
0.01
0.01
0.01
0.01
0.00
x.p(x)
0
0.081
0.092
0.072
0.056
0.07
0.054
0.042
0.048
0.036
x^2.p(x)
0
0.081
0.184
0.216
0.224
0.35
0.324
0.294
0.384
0.324
E(X) =
0.551
E(X^2) =
1.673
V(X) =
1.3694
SD(X) =
1.17021
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