Question

these data are the number of junk e - mails lena received for 9 consecutive days. 59 3 3 5 6 28 22 7 9. find the range, mean, and standard deviation. round the mean to one decimal place and the standard deviation to two decimal places, if necessary. part 1 of 3 the range is 56 e - mails. part 2 of 3 the mean is 15.8 e - mails. (round the answer to one decimal place, if necessary.) part: 2 / 3 part 3 of 3 the standard deviation is e - mails. (round the answer to two decimal places, if necessary.)

Explanation:

Step1: Recall standard - deviation formula

The formula for the sample standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$, where $x_{i}$ are the data points, $\bar{x}$ is the mean, and $n$ is the number of data points. Here $n = 9$ and $\bar{x}=15.8$.

Step2: Calculate $(x_{i}-\bar{x})^{2}$ for each data - point

For $x_1 = 59$: $(59 - 15.8)^2=(43.2)^2 = 1866.24$
For $x_2 = 3$: $(3 - 15.8)^2=(-12.8)^2 = 163.84$
For $x_3 = 3$: $(3 - 15.8)^2=(-12.8)^2 = 163.84$
For $x_4 = 5$: $(5 - 15.8)^2=(-10.8)^2 = 116.64$
For $x_5 = 6$: $(6 - 15.8)^2=(-9.8)^2 = 96.04$
For $x_6 = 28$: $(28 - 15.8)^2=(12.2)^2 = 148.84$
For $x_7 = 22$: $(22 - 15.8)^2=(6.2)^2 = 38.44$
For $x_8 = 7$: $(7 - 15.8)^2=(-8.8)^2 = 77.44$
For $x_9 = 9$: $(9 - 15.8)^2=(-6.8)^2 = 46.24$

Step3: Calculate $\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}$

$\sum_{i = 1}^{9}(x_{i}-\bar{x})^{2}=1866.24+163.84 + 163.84+116.64+96.04+148.84+38.44+77.44+46.24=2717.6$

Step4: Calculate the standard deviation

$s=\sqrt{\frac{2717.6}{9 - 1}}=\sqrt{\frac{2717.6}{8}}=\sqrt{339.7}=18.43$

Answer:

$18.43$