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exercise 1.the probabilities of laptops to have 0, 1, 2, 3 or 4 defecti…

Question

exercise 1.the probabilities of laptops to have 0, 1, 2, 3 or 4 defective parts are 0.24, 0.18, 0.26, 0.12 and 0.20, respectively. find the mean of the probability distribution.

Explanation:

Step1: Recall the formula for the mean of a probability distribution

The formula for the mean \(\mu\) of a discrete probability distribution is \(\mu=\sum_{i}x_ip_i\), where \(x_i\) are the values of the random variable and \(p_i\) are their corresponding probabilities.

Step2: Calculate each \(x_ip_i\) term

  • For \(x = 0\) and \(p=0.24\): \(0\times0.24 = 0\)
  • For \(x = 1\) and \(p = 0.18\): \(1\times0.18=0.18\)
  • For \(x = 2\) and \(p = 0.26\): \(2\times0.26 = 0.52\)
  • For \(x = 3\) and \(p = 0.12\): \(3\times0.12=0.36\)
  • For \(x = 4\) and \(p = 0.20\): \(4\times0.20 = 0.8\)

Step3: Sum up the \(x_ip_i\) terms

\(\mu=0 + 0.18+0.52 + 0.36+0.8\)

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Answer:

\(1.86\)