Q No. A company is seriously considering buying one type of machine that can save significant labor hours. The labor hour saved by the machine follows a normal distribution with mean = 2200 and a certain standard deviation. It is known that there is a fifty- fifty chance that the labor hours saved by the machine is either greater than 2400 or less than 2000. The price of the machine is Rs.860000. The incremental cost of a labor hour incurred in the company is currently Rs. 400. The company has also performed 36 trial runs to find out how the machine is fairing. The company would like to make some preliminary assessment before buying the machine. It requires your help to answer the following fill in the blank questions:
1) The standard deviation of the population distribution of the labor hours saved by the machine is ---------------
2) The standard error of the sample mean of labor hours saved is --------------
3) The probability that the sample mean of labor hours saved will exceed the break-even labor hours is ------------
To find the answers to the given questions, we need to use the information provided in the question.
1) The standard deviation of the population distribution of the labor hours saved by the machine is not directly given in the question. However, we are provided with the mean, and we know that the distribution follows a normal distribution.
Given that there is a fifty-fifty chance that the labor hours saved by the machine is either greater than 2400 or less than 2000, we can infer that the mean of the distribution is the midpoint between these two values, which is (2400 + 2000)/2 = 2200.
Since the standard deviation is not provided, we cannot directly calculate it. However, if we assume that the distribution is symmetric around the mean, we can estimate the standard deviation using the 68-95-99.7 rule of a normal distribution. According to this rule, approximately 68% of the data falls within one standard deviation of the mean, 95% falls within two standard deviations, and 99.7% falls within three standard deviations.
Assuming that the labor hours saved distribution follows this rule, we can divide the range between 2000 and 2400 into three equal parts (since it's symmetric) and approximate that each segment represents one standard deviation. Therefore, the estimated standard deviation would be (2400-2000)/3 = 133.33 (approximately).
So, the estimated standard deviation of the population distribution of the labor hours saved by the machine is approximately 133.33.
2) The standard error of the sample mean of labor hours saved represents the variability or uncertainty of the sample mean relative to the population mean. To calculate it, we need to know the sample size (n) and the standard deviation of the population (σ).
In the question, it is mentioned that 36 trial runs were performed, which represents the sample size (n = 36).
As we assumed in the previous question that the standard deviation of the population is approximately 133.33, we can use this value as an estimate of the population standard deviation (σ).
The formula for calculating the standard error of the sample mean (SE) is SE = σ / √n, where σ is the population standard deviation and √n is the square root of the sample size.
So, the standard error of the sample mean of labor hours saved is approximately 133.33 / √36 = 22.22 (approximately).
3) The break-even labor hours represent the point at which the cost of labor saved by the machine equals the cost of the machine itself. In this case, the break-even labor hours can be calculated as the cost of the machine divided by the incremental cost of a labor hour.
The cost of the machine is given as Rs. 860,000, and the incremental cost of a labor hour is given as Rs. 400.
Therefore, the break-even labor hours can be calculated as 860,000 / 400 = 2150.
To find the probability that the sample mean of labor hours saved will exceed the break-even labor hours, we need to compare the break-even labor hours with the sample mean distribution.
Since the distribution of the labor hours saved by the machine follows a normal distribution, we can use the Z-score to calculate the probability.
The Z-score is calculated as (sample mean - population mean) / (standard deviation / √n).
In this case, the population mean is 2200, the standard deviation is estimated to be 133.33, and the sample size is 36.
The Z-score for the break-even labor hours can be calculated as (2150 - 2200) / (133.33 / √36) = -7.55 (approximately).
Now, we can look up this Z-score in a standard normal distribution table or use a calculator to find the corresponding probability. The probability can be interpreted as the probability that the sample mean of labor hours saved is less than the break-even labor hours. To find the probability that the sample mean will exceed the break-even labor hours, we subtract this probability from 1.
Using a standard normal distribution table or calculator, the probability can be found as approximately 1 minus the corresponding cumulative probability for a Z-score of -7.55.
So, the probability that the sample mean of labor hours saved will exceed the break-even labor hours is approximately 1 minus the cumulative probability of -7.55 (which is nearly 0).
Please note that the exact value for the probability needs to be calculated using a standard normal distribution table or calculator, as the provided Z-score is an estimate.