Question

1. what is the sample standard deviation for the data given: 5, 10, 7, 12, 0, 20, 15, 22, 8, 2

a) 6.89
c) 10.1
b) 9
d) 7.26

1. all of the below sets have the same mean.

set 1 - standard deviation = 3.1
set 2 - standard deviation = 4.9
set 3 - standard deviation = 1.7
set 4 - standard deviation = 3.2
which set of data probably has the points closest to the mean?
a) 2
c) 3
b) 4
d) 1

1. four data sets are shown below.

set 1: {10, 19, 38, 50, 51}
set 2: {5, 21, 26, 39, 51}
set 3: {9, 38, 50, 50, 51}
set 4: {5, 28, 28, 28, 51}
which data set has the largest standard deviation?
a) set 2
c) set 4
b) set 1
d) set 3

1. how many values are within one standard deviation of the mean? 180, 313, 101, 255, 202, 198, 109, 183, 181, 113, 171, 165, 318, 145, 131, 145, 226, 113, 228, 108

a) 12
c) 11
b) 10
d) 9

Explanation:

Step1: Recall standard - deviation concept

Standard deviation measures the amount of variation or dispersion of a set of values. A smaller standard deviation indicates that the data points tend to be closer to the mean, while a larger standard deviation indicates that the data points are more spread out from the mean.

Step2: Solve question 1

1. First, find the mean $\bar{x}$ of the data set $5,10,7,12,0,20,15,22,8,2$.

• $\bar{x}=\frac{5 + 10+7+12+0+20+15+22+8+2}{10}=\frac{101}{10}=10.1$.

1. Then, find the squared - differences $(x_i-\bar{x})^2$ for each data point $x_i$.

• $(5 - 10.1)^2=(-5.1)^2 = 26.01$, $(10 - 10.1)^2=(-0.1)^2 = 0.01$, $(7 - 10.1)^2=(-3.1)^2 = 9.61$, $(12 - 10.1)^2=(1.9)^2 = 3.61$, $(0 - 10.1)^2=(-10.1)^2 = 102.01$, $(20 - 10.1)^2=(9.9)^2 = 98.01$, $(15 - 10.1)^2=(4.9)^2 = 24.01$, $(22 - 10.1)^2=(11.9)^2 = 141.61$, $(8 - 10.1)^2=(-2.1)^2 = 4.41$, $(2 - 10.1)^2=(-8.1)^2 = 65.61$.

1. Next, find the sum of squared - differences $\sum_{i = 1}^{n}(x_i-\bar{x})^2$.

• $\sum_{i = 1}^{10}(x_i-\bar{x})^2=26.01+0.01 + 9.61+3.61+102.01+98.01+24.01+141.61+4.41+65.61=474.9$.

1. Then, for sample standard deviation $s$, use the formula $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}$. Here $n = 10$, so $s=\sqrt{\frac{474.9}{9}}\approx\sqrt{52.7667}\approx6.89$.

Step3: Solve question 2

Since a smaller standard deviation means the data points are closer to the mean, among the standard - deviation values $3.1,4.9,1.7,3.2$, the set with a standard deviation of $1.7$ (Set 4) has the points closest to the mean.

Step4: Solve question 3

1. Calculate the mean for each set:

• For Set 1: $\bar{x}_1=\frac{10 + 19+38+50+51}{5}=\frac{168}{5}=33.6$.
• For Set 2: $\bar{x}_2=\frac{5+21+26+39+51}{5}=\frac{142}{5}=28.4$.
• For Set 3: $\bar{x}_3=\frac{9+38+50+50+51}{5}=\frac{198}{5}=39.6$.
• For Set 4: $\bar{x}_4=\frac{5+28+28+28+51}{5}=\frac{140}{5}=28$.

1. Calculate the squared - differences and standard deviations for each set. After calculation, we find that Set 1 has the largest standard deviation.

Answer:

1. a) 6.89
2. b) 4
3. b) Set 1