A factory has six machines. The numbers of defective chips produced by machines 1 to 6 are 1, 0, 3, 2, 1 and 0 respectively. One machine is selected at random and X denotes the number of defective chips produced by the selected machine.

Answer: Probability distribution: P(X=0)=13P(X=0)=\frac{1}{3}, P(X=1)=13P(X=1)=\frac{1}{3}, P(X=2)=16P(X=2)=\frac{1}{6}, P(X=3)=16P(X=3)=\frac{1}{6}; Mean: E(X)=76E(X) = \frac{7}{6}; Variance: Var(X)=4136\text{Var}(X) = \frac{41}{36}

Step-by-step solution

Step 1: Determine the probability distribution of X

The counts of defective chips for the 6 machines are 1, 0, 3, 2, 1, and 0. Each machine is equally likely to be selected with probability 16\frac{1}{6}. The possible values of XX are 0, 1, 2, and 3. Value 0 occurs 2 times (P(X=0)=26P(X=0) = \frac{2}{6}), value 1 occurs 2 times (P(X=1)=26P(X=1) = \frac{2}{6}), value 2 occurs 1 time (P(X=2)=16P(X=2) = \frac{1}{6}), and value 3 occurs 1 time (P(X=3)=16P(X=3) = \frac{1}{6}).

Step 2: Calculate the mean E(X)

The expected value or mean E(X)E(X) is obtained by summing the products of each value xix_i and its respective probability P(X=xi)P(X=x_i). Multiplying and adding gives 0+2+2+36=76\frac{0 + 2 + 2 + 3}{6} = \frac{7}{6}.

Step 3: Calculate E(X2X^2)

To determine the variance, we first find E(X2)E(X^2) by multiplying the square of each value by its corresponding probability. This gives 0(26)+1(26)+4(16)+9(16)=156=520\left(\frac{2}{6}\right) + 1\left(\frac{2}{6}\right) + 4\left(\frac{1}{6}\right) + 9\left(\frac{1}{6}\right) = \frac{15}{6} = \frac{5}{2}.

Step 4: Calculate the variance Var(X)

The variance is given by Var(X)=E(X2)−[E(X)]2\text{Var}(X) = E(X^2) - [E(X)]^2. Substituting the calculated values, we get 156−4936=90−4936=4136\frac{15}{6} - \frac{49}{36} = \frac{90 - 49}{36} = \frac{41}{36}.

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