Question

A sample of 100 cars driving on a freeway during a morning commute was drawn, and the number of occupants in each car was recorded The results were as follows: Occupants 1 2 3 4 5 Number of Cars 70 15 10 3 2 a. Find the sample mean number of occupants. b. Find the sample standard deviation of the number of occupants. c. Find the sample median number of occupants. d. Compute the first and third quartiles of the number of occupants. e. What proportion of cars had more than the mean number of occupants? f. For what proportion of cars was the number of occupants more than one standard deviation greater than the mean? g. For what proportion of cars was the number of occupants within one standard deviation of the mean? 1

Answer 1

Answer:

a)The required sample mean is 1.52

b)The required sample standard deviation is 0.9325

c) The median is 1 .

d) the first and third quartiles of the number of occupants is 0.15

e) Proportion of cars that had more than the mean number of occupants is 0.30

f) Proportion of cars with more than 1 standard deviation from the mean is 0.15

g)Proportion of cars within 1 standard deviation from mean  is 0.85.

Step-by-step explanation:

Step 1: a) Compute the required sample mean

The required sample mean can be obtained as:

$\begin{equation}{Mean}(\bar{x})=\frac{\sum_{i=1}^{n} f_{i} x_{i}}{\sum_{i} f_{i}} \end{equation}$

$&=\frac{152}{100} \\\\=1.52$

Step 2: b) Determine the standard deviation

The required sample standard deviation can be obtained as:

$\begin{aligned}s &=\sqrt{\frac{\sum f x^{2}}{\sum f}-\left(\frac{\sum f x}{\sum f}\right)^{2}} \\&=\sqrt{\frac{318}{100}-\left(\frac{152}{100}\right)^{2}} \\&=0.9325\end{aligned}$

Thus, the standard deviation is $0.9325.$

Step 3: c) Determine the median

The required median can be obtained as:

$\begin{aligned}\text { Median } &=\left(\frac{n+1}{2}\right) \text { th value } \\&=\left(\frac{100+1}{2}\right) \text { th value } \\&=(50.5) \text { th value } \\&=1\end{aligned}$

Thus, the median is 1 .

Step 4 d) Compute the first and third quartiles of the number of occupants

$1-0.85=0.15$

Step 5 e) what proportion of cars had more than mean number of occupants:

$\begin{aligned}&=P(X > 1.52) \\&=P(X=2)+P(X=3)+P(X=4)+P(X=5) \\&=(15+10+3+2) / 100 \\&=0.30\end{aligned}$

Step 6 f) proportion of cars more than 1 standard deviation from mean $=P(X > 1.52+0.9325)=P(X > 2.4525) \begin{aligned}&=P(X=3)+P(X=4)+P(X=5) \\&=(10+3+2) / 100 \\&=0.15\end{aligned}$

Step 7 g) proportion of cars within 1 standard deviation from mean

$P($ within 1 standard deviation from mean $)=P(1.52-0.9325 < X < 1.52+0.9325)$

$=P(0.5875 < X < 2.4525) \\=P(X=1)+P(X=2) \\=(70+15) / 100 \\=0.85$

Answer 2

Answer:

a)The required sample mean is 1.52

b)The required sample standard deviation is 0.9325

c) The median is 1 .

d) the first and third quartiles of the number of occupants is 0.15

e) Proportion of cars that had more than the mean number of occupants is 0.30

f) Proportion of cars with more than 1 standard deviation from the mean is 0.15

g)Proportion of cars within 1 standard deviation from mean  is 0.85.

Step-by-step explanation:

Step 1: a) Compute the required sample mean
The required sample mean can be obtained as:
$Mean(x) = \frac{\sum_{i=1}^{\n}f_{i}}{\sum_{i=1}^{\n}f_{i}}$
$\frac{152}{100}$$= 1.52$

Step 2: b) Determine the standard deviation
The required sample standard deviation can be obtained as:
$s =\sqrt{\frac{\sum fx^2}{\sum f}-(\frac{\sum fx}{\sum f} )^2 }$
$= \sqrt{\frac{318}{100} - (\frac{152}{100})^2 }$$= 0.9325$
Thus, the standard deviation is, $0.9325$

Step 3: c) Determine the median

The required median can be obtained as:
$Median = (\frac{n+1}{2} ) \\= (\frac{100+1}{2} )$th value

$=1$

Thus, median is $1$.

Step 4 d) Compute the first and third quartiles of the number of occupants

$1-0.85=0.15$

Step 5 e) what proportion of cars had more than mean number of occupants:
$=P(X > 1.52)\\=P(X=2)+P(X=3)+P(X=4)+P(X=5)\\=\frac{15+10+3+2}{100}\\=0.30$

Step 6 f) proportion of cars more than 1 standard deviation from mean
$= P(X > 1.52+0.9325) = P(X > 2.4525)\\=P(X=3)+P(X=4)+P(X=5)\\=\frac{(10+3+2)}{5}\\ =0.15$
Step 7 g) proportion of cars within 1 standard deviation from mean
$P($within 1 standard deviation from mean$)$

$=P(1.52-0.9325 < X < 2.4525)$

$=P(0.5875 < X < 2.4525)\\=P(X=1)+P(X=2)\\=\frac{70+15}{100}\\ =0.85$