Question

x |
P(X=x) |
|||||

0 | 0.44 | |||||

The incomplete
table at right is a discrete random variable x's
probability distribution, where x is the number of days in
a week people exercise. Answer the following: |
1 | 0.18 | ||||

2 | 0.02 | |||||

3 | ||||||

(a) | Determine the value that is missing in the table. | 4 | 0.03 | |||

5 | 0.13 | |||||

6 | 0.07 | |||||

(b) | Explain the meaning of " P(x < 2) " as it applies to the context of this problem. | 7 | 0.04 | |||

(c) | Determine the value of P(x > 4): | |||||

(d) | Find the
probability that x is at least 1. |
|||||

(e) | Find the mean (expected value) and standard deviation of this probability distribution. | |||||

Answer #1

(a) Here sum of all the probabilites shall be equal to 1.

= 1

0.44 + 0.18 + 0.02 + P(3) + 0.03 + 0.13 + 0.07 + 0.04 = 1

P(3) = 1 - 0.91 = 0.09

(b) Here the meaning of P(x < 2) is values of x that is less than 2. here it means the probability that number of days people exercise less than 2 in a week.

(c) P(x < 4) = P(0) + P(1) + P(2) + P(3) = 0.44 + 0.18 + 0.02 + 0.09 = 0.73

(d) P(x at least 1) = 1 - P(x is less than 1) =1 - P(0) = 1 - 0.44 = 0.56

(e) Mean = = 0 * 0.44 + 1 * 0.18 + 2 * 0.02 + 3 * 0.09 + 4 * 0.03 + 5 * 0.13 + 6 * 0.07 + 7 * 0.04 = 1.96

Standard deviation = Sqrt (Variance)

Variance =
- ()^{2}

= 0 * 0.44 + 1 * 1 * 0.18 + 2 * 2 * 0.02 + 3 * 3 * 0.09 + 4 * 4 * 0.03 + 5 * 5 * 0.13 + 6 * 6 * 0.07 + 7 * 7 * 0.04 = 9.28

Variance = 9.28 - 1.96 * 1.96 = 5.4384

Standard deviation = sqrt(5.4384) = 2.332

2.
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x
P(X=x)
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0
0.532
1
0.124
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0.013
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19
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