Question

eyeglassomatic manufactures eyeglasses for different retailers. the number of days it takes to fix defects in a pair of eyeglasses and the probability that it will take that number of days are in the table.
a) state the random variable.
b) if you were to make the corresponding histogram for this frequency table, what shape would the histogram be?
c) find the mean number of days to fix defects.
put the numeric answer rounded to 3 decimal places in the first box and the correct units in the second box.

Explanation:

Step1: Recall the formula for the mean of a discrete random variable

The formula for the mean \(\mu\) of a discrete random variable \(X\) is \(\mu=\sum_{x}x\cdot P(x)\), where \(x\) are the values of the random variable and \(P(x)\) are their corresponding probabilities.

Step2: Calculate each \(x\cdot P(x)\) term

• For \(x = 1\) and \(P(1)=0.235\), \(1\times0.235 = 0.235\)
• For \(x = 2\) and \(P(2)=0.112\), \(2\times0.112=0.224\)
• For \(x = 3\) and \(P(3)=0.102\), \(3\times0.102 = 0.306\)
• For \(x = 4\) and \(P(4)=0.123\), \(4\times0.123=0.492\)
• For \(x = 5\) and \(P(5)=0.133\), \(5\times0.133 = 0.665\)
• For \(x = 6\) and \(P(6)=0.114\), \(6\times0.114=0.684\)
• For \(x = 7\) and \(P(7)=0.07\), \(7\times0.07 = 0.49\)
• For \(x = 8\) and \(P(8)=0.046\), \(8\times0.046=0.368\)
• For \(x = 9\) and \(P(9)=0.019\), \(9\times0.019 = 0.171\)
• For \(x = 10\) and \(P(10)=0.013\), \(10\times0.013=0.13\)
• For \(x = 11\) and \(P(11)=0.01\), \(11\times0.01 = 0.11\)
• For \(x = 12\) and \(P(12)=0.008\), \(12\times0.008=0.096\)
• For \(x = 13\) and \(P(13)=0.006\), \(13\times0.006 = 0.078\)
• For \(x = 14\) and \(P(14)=0.004\), \(14\times0.004=0.056\)
• For \(x = 15\) and \(P(15)=0.001\), \(15\times0.001 = 0.015\)
• For \(x = 16\) and \(P(16)=0.002\), \(16\times0.002=0.032\)
• For \(x = 17\) and \(P(17)=0.001\), \(17\times0.001 = 0.017\)
• For \(x = 18\) and \(P(18)=0.001\), \(18\times0.001=0.018\)

Step3: Sum up all the \(x\cdot P(x)\) terms

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Answer:

\(4.564\) days