Why do should you not change your life policy to new one? Whether the change is due to financial reasons, change in your circumstances or simply an end to the current policy, the decision should be carefully considered before making any move. There are various questions that should be answered before taking the next step.
Comparing the two plans is helpful. For example, there is a significant difference between whole life insurance policies and term life insurance. While term life insurance is often cheaper than whole life insurance, it may not guarantee a benefit payment after expiry (Baker, 2012).
Some insurance companies require their policyholders to pay charges should they want to cancel their policy. This considerably decreases the benefit of changing plans. When planning to switch, talking to the insurance company about any fees payable is important (Fier and Andre, 2013).
Some insurance plans are modifiable should the holder’s financial needs change. It may be possible to add a new rider or reduce your life insurance coverage to better meet the policy holder’s needs (Bodie, 2013). It is important to speak to the insurance company to fully understand available options for those willing to change policies.
There are any reasons why people consider changing their life policies. However, it may not be reasonable to change the policy. Owing to the long-term nature of many life insurance policies, there is a probability that circumstances will change over the term of the policy. While it may be costly to alter a life plan, in some cases, it makes sense.
Baker, Tom eds. Embracing risk: The changing culture of insurance and responsibility. University of Chicago Press, 2010.
Bodie, Zvi. Investments. McGraw-Hill, 2013.
Fier, Stephen G., and Andre P. Liebenberg. "Life insurance lapse behavior." North American Actuarial Journal 17.2 (2013): 153-167.
- Engineers at a manufacturing plant have just been to a course to learn the basics of SPC (statistical process control). They have decided to set up their first charting application on a CNC turning center where a motor shaft is being machined. The table below represents the diameter of motor shafts. (Read down, then across: total of 100 observations)
- Find the sample median and the lower and upper quartiles (3 points)
Median = 450
Lower quartile = 441
Upper Quartile = 487
- Calculate its average, mode, standard deviation of the data set. (3 points)
Average = 450.01
Mode = 454
Standard Deviation = 13.42732
- Construct a histogram for these data. Comment on the shape of the histogram. (10 points)
- Does it resemble any of the distributions? (3 points)
The distribution of the data in the histogram resembles normal distribution. The data is clustered towards the center.
- Construct a normal probability plot of this data. (10 points)
- What are the levels of 6 Sigma? How does this level measurement process work? Please fill the table below. (5 points)
Measuring a process using Six Sigma seeks to improve a process by reducing the number of defects. Six Sigma aims to ensure continuous improvement hence facilitate customer satisfaction, profitability, and cost reduction.
|Sigma level||Defects per million opportunity||Percent defective||Percentage yield|
- Why should we reach 6 Sigma level? (5 points)
The 6 Sigma level is a level at which we are able to control a process. In the 6 sigma process, we move from defining and stating the practical problem, measuring problem by data collection, analyzing the problem through statistical procedures, improving the process through a statistical solution, and the last level which is controlling the process by converting the statistical solution to a practical solution.
- The average resistance of n=25100-ohm resistors, selected at random, is found to be x̄ = 100 ohms. The individual resistances are known to follow a normal distribution with a standard deviation of 10 ohms.
- Find the probability of a resistor to have a resistance greater than 108.9 ohms. (4 points)
= (108.9-100) / 10
= 1 – P (Z<0.89)
= 1 – 0.3133
- Find the probability of a resistor to have a resistance smaller than 88.9 ohms. (4 points)
= (88.9-100) / 10
= - 1.11
P (Z < -1.11)
- Find a 95% confidence interval for the mean resistance of all manufactured resistors. (12 points)
95% of the values in a normal distribution lie within 2 standard deviations of the mean.
Therefore, the 95% confidence interval for the mean resistance = 100 - 20 ≤ x̄ ≥ 100 + 20
80 ≤ x̄ ≥ 120
- Control charts for and R are maintained for an important quality characteristic. The sample size is n= 10, and R are computed for each sample. After 25 samples, we have found that
- Compute control limits of R chart. (7 points)
= 1.777 (36.8)
= 0.223 (36.8)
- Assuming that R chart exhibits control, estimate the standard deviation. (7 points)
σ = R / d2 (n)
d2 for a sample size of 25 = 3.931
= 36.8 / 3.931(25)
= 9. 3615
If process output is normally distributed, and if specifications are 100 ± 25, can the process meet the specifications?
UCL = 125
LCL = 75
Cp = UCL- LCL / 6 σ
= 125-75/ 6 (9.3615)
Cp < 1
Therefore, the process cannot meet the specifications.
Estimate the fraction nonconforming. (12 points)
Fraction non-conforming = P (X < 75) + P (X>125)
= P (Z < 75- 91.4 / 9.3615) + P (Z > 125- 91.4 / 9.3615
= P (Z < -1.7518) + P (Z > 3.5891)
= 0.0399 + (1-0.9998)
= 0.0399 +0.0002
Montgomery, DC, (2008). Introduction to Statistical Quality Control, Hoboken, NJ: Wiley