Solution a) The frequency distribution table for the given data has been developed in Table 1a. The group ranges show the lower and upper limits of the intervals. The interval size has been chosen as 50 hours. Table 1a: Frequency distribution table for life of bulbs Group Range Frequency < 650 0 650 - 700 2 700 - 750 1 750 – 800 1 800 – 850 4 850 – 900 12 900 – 950 13 950 - 1000 6 1000 - 1050 5 1050 - 1100 4 > 1100 0 b) The histogram for the frequency distribution of the life of the bulbs is plotted in Figure 1b. Figure 1b: Histogram for the life of bulbs c) On the basis of the histogram, it can be commented that most of the data points are concentrated towards the middle.
The frequency decreases as the group ranges move on either side of the middle groups. These properties are exhibited by a normal distribution. Hence, it can be said that the life of the bulbs follows a normal distribution. Solution 2 a) For the given data, the company would be concerned about the yearly increase in sales in each model in the month of March. Moreover, the company may also need information on the distribution of sales among various models and the change in distribution over the two years.
A simple bar graph would be sufficient for the first part while a pie chart would suffice for the second. These are shown in Figure 2a and 2b respectively. Figure 2a: Bar Graph for sales comparison over the 2 years Figure 2b: Pie Charts for sales distribution Sales Distribution March-10 Sales Distribution March-09 b) From the graphs above, it can be observed that the sales for all models of Ford have increased over the year. The percentage sales as a part of total sales for all models were similar in March 2010.
However, in March 2010, the sales are skewed towards Ranger, Explorer and Taurus. Lincoln Continental sees a big drop in percentage break-up. Solution 3 a) The arithmetic mean and median for the battery life in hours come out as 567.2 hours and 507 hours respectively. The first quartile is obtained at 407 hours while the second quartile is obtained at 761 hours ((using Excel). b) The above information would enable manufacturer to make an estimate about the expected life of the batteries.
The same can be used customer communication, advertising and promotions. c) The standard deviation for the given data is obtained as 239.8 hours. The inter-quartile range can be calculated as the difference between first and third quartiles which comes out as 354 hours. The range is computed as the difference between the maximum and minimum sample values which comes out as 843 hours. d) It can be seen that the mean battery life is 567 hours while the first quartile is obtained at 407 hours. Given these facts and considering that advertisements do require some exaggeration and self-promotion, the manufacturer is quite fair and accurate in his statement. e) The production manager could decide a cut-off value so that the guaranteed life is sufficiently high for the battery to be sold easily.
At the same time, the value should be optimal so that the monetary payment in case of default is minimal. For example, the manufacturer can select a value so that 90% of the batteries don’t need monetary pay-outs. Solution 4 a) The mean for the given sample is obtained as 24.94 minutes while the standard deviation is obtained as 230.9 minutes.
Mean is obtained as the weighted average of the middle values of waiting time ranges. The calculations for standard deviation are shown in the table 4a. Table 4a: Calculating Mean & Standard Deviation of the sample Waiting time in minutes Middle value of the range Frequency Total Waiting Time Frequency * Waiting Time 10 but less than 15 12.5 7 87.5 1093.75 15 but less than 20 17.5 14 245 4287.5 20 but less than 25 22.5 23 517.5 11643.75 25 but less than 30 27.5 13 357.5 9831.25 30 but less than 35 32.5 15 487.5 15843.75 35 but less than 40 37.5 8 300 11250 Sum 53950 Variance = Sum – Mean^2 53328.12 Mean 24.93 Standard Deviation = Variance ^. 5 230.92 b) The cumulative frequency chart for the data is shown in Figure 4a. Median = Lower Limit of interval in which middle value lies + (Total number of values/2 – Cumulative frequency of previous interval)/Frequency of interval in which middle value lies * Interval size = 20 + (49 – 27)/26 * 5 = 24.231 Figure 4a: Cumulative frequency chart c) When the waiting time is increase by 20% for all customers the mean and standard deviation are obtained as 29.9 and 277.1 respectively. Table 4.2: Calculating mean & standard deviation for increased waiting times Middle value of the range Frequency Total Waiting Time Frequency * Waiting Time 15 7 105 1575 21 14 294 6174 27 23 621 16767 33 13 429 14157 39 15 585 22815 45 8 360 16200 Sum 77688 Variance = Sum – Mean^2 76792.49 Mean 29.925 Standard Deviation = Variance ^. 5 277.1146 d) When the waiting time for all customers is increase by 20% it leads to an increase in both mean and standard deviation of the waiting time values. ` e) In order to improve service, the serving time within the restaurant can be reduced by using efficient cooking machines and increasing the number of waiters.
The seating arrangement can be changed to accommodate more customers. Solution 5 a) Simple random sampling would be the best sampling method for carrying out the survey for tourists visiting West Midlands as it would give unbiased results and ensure that the objective of the study is achieved. b) A brief survey questionnaire is shown below.
The questions are arranged in the mentioned order. i. Which age group do you belong to? 1. Less than 15 2. 15-35 3. 35-50 4. Above 50 ii. How would you rate the existing services for tourists at West Midlands? 1. Outstanding 2. Good 3. Average 4. Below Average 5. Poor iii. What fun activities would you like to include on a tour to West Midlands as a tourist? iv.
Please rate your experience of visiting West Midlands on a scale of 1 to 5? c) The survey should collect information on a tourist’s income, age, geography and gender. d) Many difficulties may be faced while conducting the survey. The tourists may simply not be interested in answering questions since they may be find it irritating on an interesting tour like this. Also, they may find it tough or cumbersome to answer open ended questions. e) Other methods of information collection which can be used are interviews, online surveys and focus groups.