Question

section 3.3 homework
due sep 8 by 11:59pm points 9 submitting an exter
section 3.3 homework
score: 4/8 answered: 6/8
question 7
approximate the sample standard deviation for the following frequ
data frequency
50 - 53 1
54 - 57 3
58 - 61 6
62 - 65 10
66 - 69 8
standard deviation =

Explanation:

Step1: Find mid - points

For the interval $50 - 53$, mid - point $x_1=\frac{50 + 53}{2}=51.5$.
For $54 - 57$, $x_2=\frac{54+57}{2}=55.5$.
For $58 - 61$, $x_3=\frac{58 + 61}{2}=59.5$.
For $62 - 65$, $x_4=\frac{62+65}{2}=63.5$.
For $66 - 69$, $x_5=\frac{66 + 69}{2}=67.5$.

Step2: Calculate the mean $\bar{x}$

$n=\sum f=1 + 3+6 + 10+8=28$.
$\sum_{i = 1}^{5}f_ix_i=1\times51.5+3\times55.5 + 6\times59.5+10\times63.5+8\times67.5$
$=51.5+166.5+357+635+540$
$=1750$.
$\bar{x}=\frac{\sum_{i = 1}^{5}f_ix_i}{n}=\frac{1750}{28}=62.5$.

Step3: Calculate $(x_i-\bar{x})^2f_i$

For $i = 1$: $(51.5 - 62.5)^2\times1=(- 11)^2\times1 = 121$.
For $i = 2$: $(55.5 - 62.5)^2\times3=(-7)^2\times3=49\times3 = 147$.
For $i = 3$: $(59.5 - 62.5)^2\times6=(-3)^2\times6=9\times6 = 54$.
For $i = 4$: $(63.5 - 62.5)^2\times10=(1)^2\times10 = 10$.
For $i = 5$: $(67.5 - 62.5)^2\times8=(5)^2\times8=25\times8 = 200$.
$\sum_{i = 1}^{5}(x_i-\bar{x})^2f_i=121+147+54+10+200=532$.

Step4: Calculate the sample standard deviation $s$

The formula for the sample standard deviation of grouped data is $s=\sqrt{\frac{\sum_{i = 1}^{k}(x_i-\bar{x})^2f_i}{n - 1}}$.
$s=\sqrt{\frac{532}{28 - 1}}=\sqrt{\frac{532}{27}}\approx4.4$.

Answer:

$4.4$