#### Jas100

Suppose a large set of islands have turtle populations that are approximately normally distributed with a mean population of 2100 per island and a standard deviation of 52.3. Suppose we take a simple random sample of 10 islands.

a) Find the probability that at least 4 of those islands from the Simple Random Sample (SRS) have a turtle population of AT LEAST 2150.
b) Find the probability that the average turtle population for the Simple Random Sample (SRS) is BETWEEN 2090 and 2110 turtles.

I HAVE NO CLUE HOW TO START OR SOLVE.

#### Archie

I would guess (not having looked at statistics for over 20 years) that you should normalise the distribution and the values 2150, 2090 and 2110 and then use the cumulative normal distribution to find for each of them P(a single island < number). From there you can use basic logic and probability to get the solutions.

#### HallsofIvy

MHF Helper
Suppose a large set of islands have turtle populations that are approximately normally distributed with a mean population of 2100 per island and a standard deviation of 52.3. Suppose we take a simple random sample of 10 islands.

a) Find the probability that at least 4 of those islands from the Simple Random Sample (SRS) have a turtle population of AT LEAST 2150.
z= (2150- 2100)/52.3= 50/52.3= 0.96 use that, and a table of the "standard normal distribution" to find the probability that the average turtle population is greater than 2150. Then use the binomial distribution with that probability to find "at least 4 out of 10".

b) Find the probability that the average turtle population for the Simple Random Sample (SRS) is BETWEEN 2090 and 2110 turtles.

I HAVE NO CLUE HOW TO START OR SOLVE.

Do you not know about the "normal distribution" and "binomial distribution"? There is a nice normal distribution app at Standard Normal Distribution Table

#### Jas100

For (A) is it correct for me to say 1-binomcdf(4, 0.16953, .3315)

For (B) is it correct for me to say normalcdf(2090, 2110, 2100, 52.3)

Please this is all that I know how to do!!!

For (A) I got 0.16953 from normalcdf(2150, 99999, 2100, 52.3)
And I got .3315 from looking at the Z table.