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these data are the number of junk e - mails lena received for 9 consecu…

Question

these data are the number of junk e - mails lena received for 9 consecutive days. 61 1 1 3 4 28 16 5 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 60 e - mails. part 2 of 3 the mean is 14.2 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 $\bar{x}=14.2$.

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

For $x_1 = 61$: $(61 - 14.2)^{2}=(46.8)^{2}=2190.24$
For $x_2 = 1$: $(1 - 14.2)^{2}=(- 13.2)^{2}=174.24$
For $x_3 = 1$: $(1 - 14.2)^{2}=(-13.2)^{2}=174.24$
For $x_4 = 3$: $(3 - 14.2)^{2}=(-11.2)^{2}=125.44$
For $x_5 = 4$: $(4 - 14.2)^{2}=(-10.2)^{2}=104.04$
For $x_6 = 28$: $(28 - 14.2)^{2}=(13.8)^{2}=190.44$
For $x_7 = 16$: $(16 - 14.2)^{2}=(1.8)^{2}=3.24$
For $x_8 = 5$: $(5 - 14.2)^{2}=(-9.2)^{2}=84.64$
For $x_9 = 9$: $(9 - 14.2)^{2}=(-5.2)^{2}=27.04$

Step3: Calculate the sum of $(x_{i}-\bar{x})^{2}$

$\sum_{i = 1}^{9}(x_{i}-\bar{x})^{2}=2190.24+174.24 + 174.24+125.44+104.04+190.44+3.24+84.64+27.04=2973.6$

Step4: Calculate the standard deviation

$s=\sqrt{\frac{2973.6}{9 - 1}}=\sqrt{\frac{2973.6}{8}}=\sqrt{371.7}=19.28$

Answer:

$19.28$