91Ó°ÊÓ

\(X\) is a random variable denoting the time (in minutes) in line spent by the customers until their turn.

Number of customers in the sample, \(n=50\)

The required sample mean is:

\(\bar{x}=\frac{1}{n}\sum_{i-1}^{n}x_{i}=\frac{275.5}{50}=5.51\)

The required sample standard deviation is:

\(s=\sqrt{\frac{1}{n-1}\sum_{i-1}^{n}(x_{i}-\bar{x})^{2}}=s=\sqrt{\frac{1}{50-1}*225.62}=2.15\)

Therefore we conclude that the sample mean is \(5.51\) minutes and the sample standard deviation is \(2.15\) minutes.

To construct a histogram, first the data is converted into a continuous frequency distribution. The frequency distribution is obtained as given below:

Using this table, the required histogram can easily be constructed by plotting the time (in minutes) on the \(x-\)axis and the number of customers on the \(y-\)axis. The histogram obtained is:

From the histogram, it can be easily seen that maximum number of customers had to wait for \(4-6\) minutes until their turn and very few customers had to wait for \(10-12\) minutes.

To construct a curve through the midpoints of the histogram, first a histogram is constructed using the continuous frequency distribution which is given as:

Using this table, the required curve is obtained by plotting the midpoints on the \(x-\)axis and the number of customers on the \(y-\)axis. The curve obtained is:

The curve shows an increase from \(0-2\) to \(4-6\) minutes and decreases thereafter. Also, it can be easily seen that maximum number of customers had to wait for \(4-6\) minutes until their turn and very few customers had to wait for \(10-12\) minutes.