Chapter 4, Problem 90CE

How long does it take to fly from Denver to Atlanta on Delta Airlines? The table below shows 56 observations on flight times (in minutes) for the first week of March 2005. (a) Use the grouped data formula to estimate the mean and standard deviation. (b) Using the ungrouped data (not shown), the ungrouped sample mean is 161.63 minutes and the ungrouped standard deviation is 8.07 minutes. How close did your grouped estimates come? (c) Why might flight times not be uniformly distributed within the second and third class intervals?

(Source: www.bts.gov.)

To determine

Estimate the mean, standard deviation from the grouped data and frequencies.

Answer to Problem 90CE

The mean, standard deviation from the grouped data and frequencies is:

Measures	Values
Mean	161.61
Standard deviation	7.93

Explanation of Solution

Calculation:

The given information is about the observation of flight times for the first week for 56 weeks.

The mean for the grouped data is:

$\overline{x}=\frac{{\displaystyle \sum _{j=1}^k{f}_j{m}_j}}{n}$

The standard deviation for the grouped data is:

$s=\sqrt{{\displaystyle \sum _{j=1}^k\frac{f_j\left(m_j-\overline{x}\right)}{n-1}}}$

Where,

$f_j$ is the frequency for j^th interval,

$m_j$ is the mid-point for j^th interval,

n is the total sample size $\left({\displaystyle \sum _{j=1}^k{f}_j}\right)$, when j lies between 1 and k.

If $x_l$ and $x_u$ are the lower and upper limits of the j^th interval, then $m_j=\frac{x_l+x_u}{2}$

The table below gives the mean for the grouped data:

From $x_l$	To $x_u$	Frequency $f_j$	Mid-point $m_j=\frac{x_l+x_u}{2}$	$f_j\times m_j$	${\left(m_j-\overline{x}\right)}^2$	$f_j{\left(m_j-\overline{x}\right)}^2$
140	150	1	$\frac{140+150}{2}=145$	145	$\begin{array}{l}{\left(145-161.61\right)}^2\\ =275.89\end{array}$	275.89
150	160	25	$\frac{150+160}{2}=155$	3875	$\begin{array}{l}{\left(155-161.61\right)}^2\\ =43.69\end{array}$	1,092.25
160	170	24	$\frac{160+170}{2}=165$	3960	$\begin{array}{l}{\left(165-161.61\right)}^2\\ =11.49\end{array}$	275.76
170	180	4	$\frac{170+180}{2}=175$	700	$\begin{array}{l}{\left(175-161.61\right)}^2\\ =179.29\end{array}$	717.16
180	190	2	$\frac{180+190}{2}=185$	370	$\begin{array}{l}{\left(185-161.61\right)}^2\\ =547.09\end{array}$	1,094.18
		${\displaystyle \sum f_j}=56$		$\begin{array}{l}{\displaystyle \sum _{j=1}^5{f}_j{m}_j}\\ =9,050\end{array}$		$\begin{array}{l}{\displaystyle \sum _{j=1}^5{f}_j{\left(m_j-\overline{x}\right)}^2}\\ =3,455.24\end{array}$

The mean for the grouped data is:

Substitute the values ${\displaystyle \sum f_j}=56$ and ${\displaystyle \sum _{j=1}^5{f}_j{m}_j}=9,050$ to get the mean,

$\begin{array}{c}\overline{x}=\frac{9,050}{56}\\ =161.61\end{array}$

Thus, the mean from the grouped data and frequencies is 161.61.

The standard deviation for the grouped data is:

Substitute the values ${\displaystyle \sum f_j}=56$ and ${\displaystyle \sum _{j=1}^5{f}_j{\left(m_j-\overline{x}\right)}^2}=3,455.24$ to get the standard deviation,

$\begin{array}{c}s=\sqrt{\frac{3,455.24}{56-1}}\\ =\sqrt{\frac{3,455.24}{55}}\\ =\sqrt{62.83}\\ =7.93\end{array}$

Thus, the standard deviation from the grouped data and frequencies is 7.93.

To determine

Explain how the value of grouped estimate appeared closer to the ungrouped estimate.

Explanation of Solution

It is given that the value of ungrouped sample mean is 161.63 and ungrouped standard deviation is 8.07 minutes.

From the previous part (a), the value of grouped sample mean is 161.61 and grouped standard deviation is 7.93 minutes.

By the observing the value of grouped observation, the measures appear little bit nearer to ungrouped values.

To determine

Explain the reason that the flight times for the second and third interval is not distributed uniformly.

Explanation of Solution

A careful inspection on the observation states that the distribution of the flight timings within the class interval 150-160 may be skewed to right with many more observation appear nearer to 160 and the distribution of the flight timings within the class interval 160-170 may be skewed to left with many more observation appear nearer to 160. The observation has a peculiar shape in its distribution. Thus, the shape of the distribution will be like slightly normal or somewhat skewed to right.