QUESTION IMAGE
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 19.28 e - mails. (round the answer to two decimal places, if necessary.) try again
Step1: Calculate the range
The range is the difference between the maximum and minimum values. The maximum value in the data set {61, 1, 1, 1, 3, 4, 28, 16, 5, 9} is 61 and the minimum is 1. So, $Range=61 - 1=60$.
Step2: Calculate the mean
The mean $\bar{x}$ of a data - set $x_1,x_2,\cdots,x_n$ is given by $\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}$. Here, $n = 9$, and $\sum_{i=1}^{9}x_i=61 + 1+1+1+3+4+28+16+5+9 = 129$. Then $\bar{x}=\frac{129}{9}\approx14.3$.
Step3: Calculate the standard deviation
The formula for the sample standard deviation $s$ is $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}$.
First, calculate $(x_i-\bar{x})^2$ for each $x_i$:
- For $x_1 = 61$: $(61 - 14.3)^2=(46.7)^2 = 2180.89$
- For $x_2 = 1$: $(1 - 14.3)^2=(- 13.3)^2 = 176.89$
- For $x_3 = 1$: $(1 - 14.3)^2=(-13.3)^2 = 176.89$
- For $x_4 = 1$: $(1 - 14.3)^2=(-13.3)^2 = 176.89$
- For $x_5 = 3$: $(3 - 14.3)^2=(-11.3)^2 = 127.69$
- For $x_6 = 4$: $(4 - 14.3)^2=(-10.3)^2 = 106.09$
- For $x_7 = 28$: $(28 - 14.3)^2=(13.7)^2 = 187.69$
- For $x_8 = 16$: $(16 - 14.3)^2=(1.7)^2 = 2.89$
- For $x_9 = 5$: $(5 - 14.3)^2=(-9.3)^2 = 86.49$
- For $x_{10}=9$: $(9 - 14.3)^2=(-5.3)^2 = 28.09$
The sum $\sum_{i = 1}^{9}(x_i - 14.3)^2=2180.89+176.89+176.89+176.89+127.69+106.09+187.69+2.89+86.49+28.09 = 3250.1$
Then $s=\sqrt{\frac{3250.1}{9 - 1}}=\sqrt{\frac{3250.1}{8}}\approx20.16$
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Part 1 of 3: The range is 60 e - mails.
Part 2 of 3: The mean is 14.3 e - mails.
Part 3 of 3: The standard deviation is 20.16 e - mails.