A company that sells annuities must base the annual payout on the
probability distribution of the length of life of the participants in the plan. Suppose the probability distribution of the lifetimes of the participants is approximately a normal distribution with a mean of 68 years and standard deviation of 3.5 years. What proportion of the plan receipts would receive payments beyond age 75?
A. 0.1949 B. 0.0228 C. 0.5134 D. 0.7768
There are 2,000 eligible voters in a precinct. Despite protests from Knowledgeable persons that a sample size of 500 was too large in relation to the total, the 500 selected at random were asked to indicate whether they planned to vote for the Democratic incumbent or the Republican challenger. Of the 500 surveyed, 350 said they were going to vote for the Republican incumbent. Using a 0.99 confidence level, what are the confidence limits for the proportion that plan to vote the Republican incumbent?
A. 0.826 and 0.926 B. 0.397 and 0.797
Merlyn
March 9, 2010 at 7:05 pm
For any normal random variable X with mean μ and standard deviation σ, X ~ Normal(μ, σ)
you can translate into standard normal units by:
Z = (X – μ) / σ
where Z ~ Normal(μ = 0, σ = 1). You can then use the standard normal cdf tables to get probabilities.
If you are looking at the mean of a sample, then remember that for any sample with a large enough sample size the mean will be normally distributed. This is called the Central limit theorem.
If a sample of size is is drawn from a population with mean μ and standard deviation σ then the sample average xBar is normally distributed
with mean μ and standard deviation σ / √(n)
———-
applet for finding the values
http://www-stat.stanford.edu/~naras/jsm/FindProbability.html
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question 1
Let X be the lifetime of the participants
X ~ Normal(μ = 68, σ = 3.5)
find P( X > 75 )
= P( Z > (75 – 68) / 3.5)
= P( Z > 2 )
= 0.02275013
————
question 2
large sample confidence intervals are used to find a region in which we are 100 (1-α)% confident the true value of the parameter is in the interval.
For large sample confidence intervals for the proportion in this situation you have:
pHat ± z * sqrt( (pHat * (1-pHat)) / n)
where pHat is the sample proportion
z is the zscore for having α% of the data in the tails, i.e., P( |Z| > z) = α
n is the sample size
the critical point is P( Z > z) = 0.005
z = 2.575829
the CI is
350/500 ± 2.575829 * sqrt (350 / 500 * (1 – 350/500) / 500)
= (0.6472112, 0.7527888)