Question

The average life of individual is 70 years. With a standard deviation of 5.5 years. Assume that the lives of these individuals is normally distributed. a. Find the probability that a mean life of a random sample of 5 such turtles falls between 60 and 80 years. b. Find the mean data value that separates the top 10% from the rest of the means computed from a random sample of size 5.

Answer 1

Answer:

The answer is below

Step-by-step explanation:

Given that:

mean (μ) = 70 years, standard deviation (σ)= 5.5 years.

a) The z score measures how many standard deviation a raw score is above or below the mean. It is given as:

$z=\frac{x-\mu}{\sigma}$, for a sample size of n, the z score is: $z=\frac{x-\mu}{\sigma/\sqrt{n} }$

Given a sample of 5 turtles, we have to calculate the z score for x = 60 and x = 80.

For x = 60:

$z=\frac{x-\mu}{\sigma/\sqrt{n} }=\frac{60-70}{5.5/\sqrt{5} } =-4.07$

For x = 80:

$z=\frac{x-\mu}{\sigma/\sqrt{n} }=\frac{80-70}{5.5/\sqrt{5} } =4.07$

The probability that a mean life of a random sample of 5 such turtles falls between 60 and 80 years = P(60 < x < 80) = P(-4.07 < z < 4.07) = P(z < 4.07) - P(z < -4.07) = 1 - 0 = 1 = 100%

b) The z score that corresponds to top 10% is -1.28.

$z=\frac{x-\mu}{\sigma/\sqrt{n} }\\\\-1.28=\frac{x-70}{5.5/\sqrt{5} }\\ x-70=-3\\x=70-3\\x=67\ years$