Question

3.17 A bank branch located in a commercial district of a city has the business objective of developing an improved process for serving customers during the noon-to-1:00 P.M. lunch period. The waiting time, in minutes, is defined as the time the customer enters the line to when he or she reaches the teller window. Data col- lected from a sample of 15 customers during this hour are stored in Bank1 4.21 5.55 3.02 5.13 4.77 2.34 3.54 3.20 .50 6.10 0.38 5.12 6.46 6.19 3.79 a. Compute the mean and median. b. Compute the variance, standard deviation, range, coefficient of c. Are the data skewed? If so, how? d. As a customer walks into the branch office during the lunch variation, and Z scores. Are there any outliers? Explain. hour, she asks the branch manager how long she can expect to wait. The branch manager replies, "Almost certainly less than five minutes." On the basis of the results of (a) through (c). evaluate the accuracy of this statement.

Answer 1

Based on the given data points,

a. = 4.29

After sorting the data, the middle no. of the data, the 7th number is the median.

Median = 4.5

b. = 2.68

Standard deviation is computed by taking the square root of variance.

Standard deviation$=sqrt{2.68}$=  1.64

Range is nothing but the difference between the largest and the smallest of the observations.

Range= 6.46-0.38= 6.08

Co-efficient of Variation = Standard deviation / Mean * 100

= (1.64 / 4.29) * 100

= 38%

Z scores can be computed using the formula:

To identify the outliers we first need to compute the inter-quartile range (IQR) which is the difference between the 3rd and 1st quartile.

Using excel function,

OX QUARTILE(A2:A16,]1 AVERAGE 1 ObservationsZ score 4.21 5.55 3.02 5.13 4.77 2.34 3.54 3.2 4.5 0.047 -QUARTILE(A2:A16,1 QUARTILE(array, quart) 4 0.773 0.515 0.295 1.188 0.456 0.663 0.130 1.107 2.385 0.509 1.327 1.162 0.303 10 12 13 14 15 16 0.38 5.12 6.46 6.19 3.79

We get, Q₁ = 3.37 and Q₃ = 5.34

Hence, IQR = 5.34 - 3.37 = 1.97

Any points that lie outside the range i.e less than( Q₁ - 1.5*IQR ) and greater than (Q₃ + 1.5*IQR ) are considered to be outliers in the data.

Here, Q₁ - 1.5*IQR = 3.37 - 1.5 * 1.97 = 0.415

We have one observation with value less than 0.415 (0.38).Hence it is considered an outlier.

Also, Q₃ + 1.5*IQR = 5.34 + 1.5 * 1.97 = 8.295.

No observation has value greater than 8.295.

Hence, we have one outlier in the dataset.

c. From the histogram of the data,

The data appears to be moderately skewed, with a longer left tail.

Using the excel function 'SKEW', we find that, skew = -0.833.

A negative skew supports our conclusion.

d. According to the claim, wait time less than 5 = less than Z =

From the above histogram, we find that majority of Z scores are greater than or equal to 0.433. Hence the chances of the score being otherwise is comparatively lower.

We do not have sufficient evidence to support the claim based on this sample data with certainty.