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.

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:

, for a sample size of n, the z score is:

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

For x = 60:

For x = 80:

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.

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

he can buy 4lbs. of live bait and 1lbs of natural bait

step-by-step explanation: