Answer
a. 0.856
b. 0.78071
c. It is not unusual
d. 13.65 years old
Step-by-step explanation:
The life span of a domestic cat is normally distributed with a mean of 15.7 years and a standard deviation of 1.6 years.
We solve this question using z score formula:
z = (x-μ)/σ, where
x is the raw score
μ is the population mean
σ is the population standard deviation.
a. Find the probability that a cat will live to be older than 14 years.
For x > 14 years
z = 14 - 15.7/1.6
z = -1.0625
Probability value from Z-Table:
P(x<14) = 0.144
P(x>14) = 1 - P(x<14) = 0.856
b. Find the probability that a cat will live between 14 and 18 years.
For x = 14 years
z = 14 - 15.7/1.6
z = -1.0625
Probability value from Z-Table:
P(x = 14) = 0.144
For x = 18 years
z = 18 - 15.7/1.6
z= 1.4375
Probability value from Z-Table:
P(x = 18) = 0.92471
The probability that a cat will live between 14 and 18 years is calculated as:
P(x = 18) - P(x = 14)
0.92471 - 0.144
= 0.78071
c. If a cat lives to be over 18 years, would that be unusual? Why or why not?
For x > 18 years
z = 18 - 15.7/1.6
z= 1.4375
Probability value from Z-Table:
P(x<18) = 0.92471
P(x>18) = 1 - P(x<18) = 0.075288
Converting this to percentage:
0.075288 × 100 = 7.5288%
Hence, 7.5288% of the cats live to be over 18 years. Hence, it is not unusual.
d. How old would a cat have to be to be older than 90% of other cats?
From the question above, 10% of the cats would be older than 90% of other cats.
Hence, we find the z score of the 10th percentile
= -1.282
Hence,
-1.282 = x - 15.7/1.6
Cross Multiply
-1.282 × 1.6 = x - 15.7
- 2.0512 = x - 15.7
x = 15.7 -2.0512
x = 13.6488 years old
Approximately = 13.65 years old
