4. light bulb lifetime. find the mean and median for the data in the following table.\nlife (hours) of 50…

Explanation:

Step1: Find mid - points of intervals

For interval $799.5 - 899.5$, mid - point $x_1=\frac{799.5 + 899.5}{2}=849.5$. For interval $899.5 - 999.5$, mid - point $x_2=\frac{899.5+999.5}{2}=949.5$. For interval $999.5 - 1099.5$, mid - point $x_3=\frac{999.5 + 1099.5}{2}=1049.5$. For interval $1099.5 - 1199.5$, mid - point $x_4=\frac{1099.5+1199.5}{2}=1149.5$. For interval $1199.5 - 1299.5$, mid - point $x_5=\frac{1199.5+1299.5}{2}=1249.5$.

Step2: Calculate the mean

The formula for the mean $\bar{x}=\frac{\sum_{i = 1}^{n}f_ix_i}{\sum_{i=1}^{n}f_i}$. $\sum_{i = 1}^{n}f_i=3 + 8+21 + 10+8=50$. $\sum_{i = 1}^{n}f_ix_i=3\times849.5+8\times949.5+21\times1049.5+10\times1149.5+8\times1249.5$ $=2548.5+7596+22039.5+11495+9996$ $=53675$. $\bar{x}=\frac{53675}{50}=1073.50$.

Step3: Calculate the median

Since $n = 50$ (even), the median class is the class that contains the $\frac{n}{2}=25$th and $\frac{n}{2}+1 = 26$th ordered observations. The cumulative frequencies are: For the first class: $F_1 = 3$. For the second class: $F_2=3 + 8=11$. For the third class: $F_3=3 + 8+21=32$. The median class is $999.5 - 1099.5$. $L = 999.5$ (lower limit of median class), $n = 50$, $F = 11$ (cumulative frequency of the class before the median class), $f = 21$ (frequency of the median class), $h=100$ (class width). The formula for the median $M = L+\frac{\frac{n}{2}-F}{f}\times h$. $M=999.5+\frac{25 - 11}{21}\times100$ $=999.5+\frac{14}{21}\times100$ $=999.5+\frac{1400}{21}\approx999.5 + 66.67=1066.17$.

Answer:

The mean is $1073.50$. The median is $1066.17$.