Question

In: Statistics and Probability

x P(x) 0 0.797 1 0.081 2 0.049 3 0.025 4 0.014 5 0.011 6 0.008...

x P(x)

0 0.797

1 0.081

2 0.049

3 0.025

4 0.014

5 0.011

6 0.008

7 0.007

8 0.005

9 0.003

Does the given information describe a probability? distribution?

Assuming that a probability distribution is? described, find its mean and standard deviation.

The mean is?

The standard deviation is?

The maximum usual value is?

The minimum usual value is?

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 0.122?, which is greater than 0.05.

B. ?No, because the probability of having 1 bumper sticker is 0.081?, which is greater than 0.05.

C. ?Yes, because the probabilities for random variable x from 2 to 9 are all less than 0.05.

D. Not enough information is given.

Expert Solution

The sum of probabilities P(x) is,

0.797 + 0.081 + 0.049 + 0.025 + 0.014 + 0.011 + 0.008 + 0.007 + 0.005 + 0.003 = 1

As, the sum of probabilities is 1, the given information describe a probability distribution.

Mean = E(x) =

= 0 * 0.797 + 1 * 0.081 + 2 * 0.049 + 3 * 0.025 + 4 * 0.014 + 5 * 0.011 + 6 * 0.008 + 7 * 0.007 + 8 * 0.005 + 9 * 0.003 = 0.529

E(x²) =

= 0² * 0.797 + 1² * 0.081 + 2² * 0.049 + 3² * 0.025 + 4² * 0.014 + 5² * 0.011 + 6² * 0.008 + 7² * 0.007 + 8² * 0.005 + 9² * 0.003 = 2.195

Variance = E(x²) - E(x)² = 2.195 - 0.529² = 1.915159

Standard deviation = = 1.383893

Maximum usual value is 9

Minimum usual value is 0

The probability of car to have more than one bumper = P(X > 1)

= 1 - P(X 1)

= 1 - [P(X=0) + P(X = 1)]

= 1 - (0.797 + 0.081)

= 0.122

As, the probability of car to have more than one bumper is greater than 0.05, it is not unusual. Thus, the correct option is,

A. No, because the probability of more than 1 bumper sticker is 0.122?, which is greater than 0.05.

orchestra answered 2 years ago

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