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.
x P(x)
0 0.800
1 0.088
2 0.045
3 0.017
4 0.014
5 0.012
6 0.009
7 0.007
8 0.006
9 0.002
A. Does the given information describe a probability distribution?
B. Assuming that a probability distribution is described, find its mean and standard deviation.
C. Use the range rule of thumb to identify the range of values for usual numbers of bumper stickers.
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 0.119, which is greater than 0.5.
b. Yes, because the probabilities for random variables x from 2 to 9 are all less than 0.05.
c. No, because the probability of having 1 bumper sticker is 0.099, which is greater than 0.05.
d. Not enough information is given.
a)Yes as sum of all probability is equal to 1 and individual probability is between 0 and 1
b)
x | f(x) | xP(x) | x2P(x) |
0 | 0.800 | 0.000 | 0.000 |
1 | 0.088 | 0.088 | 0.088 |
2 | 0.045 | 0.090 | 0.180 |
3 | 0.017 | 0.051 | 0.153 |
4 | 0.014 | 0.056 | 0.224 |
5 | 0.012 | 0.060 | 0.300 |
6 | 0.009 | 0.054 | 0.324 |
7 | 0.007 | 0.049 | 0.343 |
7 | 0.006 | 0.042 | 0.294 |
9 | 0.002 | 0.018 | 0.162 |
total | 0.508 | 2.068 | |
E(x) =μ= | ΣxP(x) = | 0.5080 | |
E(x2) = | Σx2P(x) = | 2.0680 | |
Var(x)=σ2 = | E(x2)-(E(x))2= | 1.8099 | |
std deviation= | σ= √σ2 = | 1.3453 |
from above mean =0.508
and std deviation =1.3453
c)
minimum usual value =μ-2*σ = | -2.183 | ||
maximum usual value =μ+2*σ = | 3.199 |
Based on data from a car bumper sticker study, when a car is randomly selected, the...
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