Exam 1: Overview and Descriptive Statistics

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a. From this frequency distribution, the proportion of wafers that contained at least two particles is (100-1-2)/100 = .97, or 97%. In a similar fashion, the proportion containing at least 6 particles is (100 - 1-2-3-12-11-15)/100 = 56/100 = .56, or 56%.
b. The proportion containing between 4 and 9 particles inclusive is (11+15+18+10+12+4)/100 = 70/100 = .70, or 70%. The proportion that contain strictly between 4 and 9 (meaning strictly more than 4 and strictly less than 9) is (15+ 18+10+12)/100= 55/100 = .55, or 55%.

Correct Answer:

Verified

a. $$\begin{array} { c c c } \hline \text { Number Nonconforming } & \text { Frequency } & \text { Relative Frequency } \\\hline 0 & 7 & 0.117 \\\hline 1 & 12 & 0.200 \\\hline 2 & 13 & 0.217 \\\hline 3 & 14 & 0.233 \\\hline 4 & 6 & 0.100 \\\hline 5 & 3 & 0.050 \\\hline 6 & 3 & 0.050 \\\hline 7 & 1 & 0.017 \\\hline 8 & 1 & 0.017 \\\hline\end{array}$$ 1.001
The relative frequencies don't add up exactly to 1because they have been rounded
b. The number of batches with at most 4 nonconforming items is 7+12+13+14+6=52, which is a proportion of 52/60=.867. The proportion of batches with (strictly) fewer than 4 nonconforming items is 46/60=.767.

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a. $\bar { y } = \frac { \sum y _ { i } } { n } = \frac { \sum \left( x _ { i } + c \right) } { n } = \frac { \sum x _ { i } } { n } + \frac { n c } { n } = \bar { x } + c$ $\tilde{y}$ = the median of $\left( x _ { 1 } + c , x _ { 2 } + c , \ldots \ldots , x _ { n } + c \right)$ = median of $\left( x _ { 1 } , x _ { 2 } , \ldots , x _ { n } \right) + c = \tilde{x} + c$ b. $\bar { y } = \frac { \sum y _ { i } } { n } = \frac { \sum \left( x _ { i } \cdot c \right) } { n } = \frac { c \sum x _ { i } } { n } = c \bar { x }$ $\tilde{y}$ = $\left( c x _ { 1 } , c x _ { 2 } , \ldots \ldots , c x _ { n } \right) = c \cdot \operatorname { median } \left( x _ { 1 } , x _ { 2 } , \ldots \ldots , x _ { n } \right) = c \bar{x}$