Answer:
a) 0.9920 = 99.20% probability that 15 or more will live beyond their 90th birthday
b) 0.2946 = 29.46% probability that 30 or more will live beyond their 90th birthday
c) 0.6273 = 62.73% probability that between 25 and 35 will live beyond their 90th birthday
d) 0.0034 = 0.34% probability that more than 40 will live beyond their 90th birthday
Step-by-step explanation:
We solve this question using the normal approximation to the binomial distribution.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
The standard deviation of the binomial distribution is:
Normal probability distribution
Problems of normally distributed distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that ,
.
In this problem, we have that:
Sample of 723, 3.7% will live past their 90th birthday.
This means that .
So for the approximation, we will have:
(a) 15 or more will live beyond their 90th birthday
This is, using continuity correction, , which is 1 subtracted by the pvalue of Z when X = 14.5. So
has a pvalue of 0.0080
1 - 0.0080 = 0.9920
0.9920 = 99.20% probability that 15 or more will live beyond their 90th birthday
(b) 30 or more will live beyond their 90th birthday
This is, using continuity correction, , which is 1 subtracted by the pvalue of Z when X = 29.5. So
has a pvalue of 0.7054
1 - 0.7054 = 0.2946
0.2946 = 29.46% probability that 30 or more will live beyond their 90th birthday
(c) between 25 and 35 will live beyond their 90th birthday
This is, using continuity correction, , which is the pvalue of Z when X = 35.5 subtracted by the pvalue of Z when X = 24.5. So
X = 35.5
has a pvalue of 0.9573
X = 24.5
has a pvalue of 0.3300
0.9573 - 0.3300 = 0.6273
0.6273 = 62.73% probability that between 25 and 35 will live beyond their 90th birthday.
(d) more than 40 will live beyond their 90th birthday
This is, using continuity correction, P(X > 40+0.5) = P(X > 40.5), which is 1 subtracted by the pvalue of Z when X = 40.5. So
has a pvalue of 0.9966
1 - 0.9966 = 0.0034
0.0034 = 0.34% probability that more than 40 will live beyond their 90th birthday