Question

A survey is carried out to study the number of hours, X, per day spent on using the Internet by the customers of an Internet Service Provider (ISP). Responses from 15 randomly selected customers give the following data:

3 2 5 6 1 5 3 4 5 2 4 5 9 5 1

(a) Determine the mean and variance of the data set. 

(b) Is there an outlier in the data set? Justify your answer. 

(c) Suppose a selected customer is found to spend at most 5 hours on using the Internet. What is the

probability that this customer spends more than 3 hours on using the Internet?

(d) Suppose the customer service manager of the ISP discovers that one data point has been mistakenly recorded to 4, and that data point should be less than 4. State whether the mode, median and mean of the correct sample become smaller than, greater than or the same as those of the incorrect sample.

Explain your answer.

Answer 1

Consider a random variable X representing a number of hours per day spend on using the Internet by the customers. The 15 randomly selected customers give the following values:

X : 3, 2, 5, 6, 1, 5, 3, 4, 5, 2, 4, 5, 9, 5 and 1.

(a):

The mean of a set of observations is the sum of the observations divided by the total number of observations. It is denoted by x̅. Mathematically, it is defined as,

where  ∑x_i denotes the sum of all data points and n denotes the total number of observations.

• As there are 15 records thus the value of n is 15.

• Add all the observations and divide it by the total number of observations.

• Substitute the respective values to obtain the value of x̅.

Thus, the mean of the given data set is 4.

The sample standard deviation indicates how far are the data points from the mean value in the sample and is denoted by s. The square value of s i.e. s² is the variance of the data set.

Mathematically, it is defined as,

, where ∑x_i²

are the sum of the square of data points, x̅ is the mean, and 'n' is the total number of observations.

Substitute the respective values to obtain the value of s².

Thus, the value of variance, s² is 4.429.

(b):

Box plot also known as box-and-whisker diagrams represents the distribution of data based on the five measures like minima, maxima, first quartile, median and third quartile of the data.

• It contains a box whose bottom side represents the first quartile and the top side represents the third quartile.

• The desired box plot is shown below of the given observations.

Minitab Technology: Graph-> Boxplot-> Select desired variable under graph variables-> OK.

The obtained Box Plot is:

The above box plot suggests that there is no outlier in the given data (Hours).

The distribution of hours is skewed as it has long whisker on the right side.

• The five measure summary of the box plot listed below:

Minimum	1
Q1	2
Median	4
Q3	5
Maximum	9

(c):

As the continuous random variable X denotes the rate i.e. hours per day thus it follows the exponential distribution with parameter λ per hour in a day.

Since the mean value of X is 4 hours thus the value of λ will be 1/4 per hour. Then, the pdf of X will be,

Thus, the probability that a customer spends more than 3 hours given that he spends at most 5 hours on the using the internet i.e..

Obtian the value of $\begin{aligned}P(3< X \leq 5) \end{aligned}$ and respectively.

and

$\begin{aligned}P(X \leq 5)&=1-e^{-5/4}\\&=1-0.287\\&=0.713\end{aligned}$

Divide 0.185 and 0.713 to obtain the desired probability.

That is,

$\begin{aligned}P(X> 3|X\leq 5)&=\frac{P(3< X \leq 5)}{P(X\leq 5)}\\&=\frac{0.185}{0.713}\\&=0.259 \end{aligned}$

Thus, the value of desired probability is 0.259.

(d):

Mode is the most frequent value of the data, thus the value of mode is 5 as it occurs most times as compared to others.

• The value of both, mean and median is 4.

• If the record 4 of the data was incorrect and it was actually less than 4 then the value of mean and median will affect as both depend on incorrect record i.e. 4.

• Since the actual record is less than 4 thus the value of mean and median will be less than 4 (or less than the mean and median based on the incorrect sample).

• As the mode of the given data is 5 which is not based on the incorrect record thus the value of the model will remain unchanged.