SOLUTION: For this discussion you will post a problem that identifies a variable that might have a normal distribution. Then pose two questions. The first asks for a probability and the se

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Question 1200085: For this discussion you will post a problem that identifies a variable that might have a normal distribution. Then pose two questions. The first asks for a probability and the second gives either a quartile or percentile and asks to find the value of x that has that quartile or percentile. Be sure that in your initial post you include the mean and standard deviation. You can quickly look up the mean and to find the standard deviation, if it is not given note that 95% of the data is within two standard deviations from the mean. For example if the mean is 15 and based on common knowledge, the vast majority lies between 9 and 21, then 21 is 6 from 15, half of 6 is 3, thus the standard deviation is around 3.
Below is what I have. I feel like I am making this more complicated than I need to, But I'm not sure
The average sea turtle life span has a normal distribution of 50 years and a standard deviation of 5 years.
1. Find the probability that a sea turtle lives 40 years
2. Find the IQR.
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
The average sea turtle life span has a normal distribution of 50 years and a standard deviation of 5 years.

1. Find the probability that a sea turtle lives more than 40 years

mean is 50 and standard deviation is 5.

using the z-score calculator at https://davidmlane.com/hyperstat/z_table.html, the probability that the turtle lives more than 40 years is equal to .9772.

2. Find the IQR.

i believe the Interquartile range is between 25% and 75%, otherwise known as the middle 50% of your values.
using the same z-score calculator used above, the probability of the life of the turtle being in the inter-quartile range is equal to between 46.628 and 53.372.

if you work from the z-score, you should get the same answer.

1. Find the probability that a sea turtle lives more than 40 years

z = (x - m) / s
z is the z-score
x is the raw score
m is the raw mean
s is the standard deviation

formula becomes z = (40 - 50) / 5 = -10 / 5 = -2.
area to the left of a z-score of -2 is equal to .0228.
area to the right of that z-score is equal to 1 - .0228 = .9772

2. Find the IQR.

you need to find the z-score that has .25 of the area to the left of it and you need to find the z-score that has .75 of the area to the left of it.
the area in between is equal to the larger area minus the smaller area.
z-score that has .25 of the area to the left of it is equal to -.674
z-score that has .75 of the area to the left of it is equal to .674
area in between is .674 minus

when z-score = -.674, the z-score formula becomes -.674 = (x - 50) / 5.
solve for x to get x = -.674 * 5 + 50 = 46.63.
when z-score = .674, the z-score formula becomes .674 = (x - 50) / 5.
solve for x to get x = .674 * 5 + 50 = 53.37.
the difference between you got using the raw scores and using the z-scores is do to the valculator rounding.
using my t-84 plus calculator, i got.
.25 area to the left of z-score = -.6744897495.
raw score = 46.628
.75 area to the left of z-score = .6744897495.
raw score = 53.372