Respuesta :
Answer:
15.87% probability of its life expectancy being greater than seven years
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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.
In this problem, we have that:
Life expectancy of 5 years and 6 months
6 months is half an year, so [tex]\mu = 5.5[/tex]
Standard deviation of 1 year and 6 months.
So [tex]\sigma = 1.5[/tex]
Probability of its life expectancy being greater than seven years
This is 1 subtracted by the pvalue of Z when X = 7.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{7 - 5.5}{1.5}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a pvalue of 0.8413
1 - 0.8413 = 0.1587
15.87% probability of its life expectancy being greater than seven years
