The following data are based on a survey taken by a consumer research firm. X = number of internet connected devices in a household and P(X) = probability of U.S. households. What value for P(X = 2 ) will make this a PDF?

X 0 1 2 3 4 5
P(X) 0.03 0.11 ? 0.39 0.12 0.07

Question

The following data are based on a survey taken by a consumer research firm. X = number of internet connected devices in a household and P(X) = probability of U.S. households. What value for P(X = 2 ) will make this a PDF?

X	0	1	2	3	4	5
P(X)	0.03	0.11	?	0.39	0.12	0.07

BlackTom AI · Accepted Answer

A discrete probability distribution must have all probabilities nonnegative and must sum to 1. Use that constraint to find the missing probability P(X = 2).

1) Relevant concepts:
- For a discrete random variable X, the probabilities P(X = x_i) must satisfy Σ P(X = x_i) = 1.
- Each individual probability must lie between 0 and 1 inclusive.

2) Step-by-step reasoning and calculations:
- List the known probabilities:
  P(X = 0) = 0.03
  P(X = 1) = 0.11
  P(X = 2) = ? (call this p)
  P(X = 3) = 0.39
  P(X = 4) = 0.12
  P(X = 5) = 0.07
- Sum the known probabilities excluding p:
  0.03 + 0.11 + 0.39 + 0.12 + 0.07
- Compute the intermediate sums:
  0.03 + 0.11 = 0.14
  0.14 + 0.39 = 0.53
  0.53 + 0.12 = 0.65
  0.65 + 0.07 = 0.72
- Use the total-sum constraint:
  p + 0.72 = 1
- Solve for p:
  p = 1 - 0.72
  p = 0.28

3) Check validity:
- 0 ≤ 0.28 ≤ 1, so this value is a valid probability.
- Verify the full sum: 0.03 + 0.11 + 0.28 + 0.39 + 0.12 + 0.07 = 1.00.

4) Why this is correct:
- This value is the unique number that makes the probabilities add to 1, satisfying the defining property of a probability distribution. Any other value would either make the total less than 1 (if smaller) or greater than 1 (if larger), violating the requirement. Negative or >1 values would also be invalid.

5) Other possible values:
- No alternative numeric choices are valid; only 0.28 makes the distribution a proper PDF and lies within [0,1].

Conclusion: The value that makes this a valid probability distribution is P(X = 2) = 0.28, matching the provided answer.

The following data are based on a survey taken by a consumer research firm. X = number of internet connected devices in a household and P(X) = probability of U.S. households. What value for P(X = 2 ) will make this a PDF? X 0 1 2 3 4 5 P(X) 0.03 0.11 ? 0.39 0.12 0.07数值题

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