Okay, so I am currently learning about normal distribution, and I don't know how to do the question that looks like this (I'll try my best to convey it, but I don't have all the right symbols on my laptop): P(X is less than or equal to 64)
The mean is 76 and the standard deviation is 6.
- Please no vague or unsure answers.
Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
x = 64 is the given raw score
the first step is to convert this to a standard z-score
z = (x - mu)/sigma
z = (x - 76)/6
z = (64 - 76)/6
z = -12/6
z = -2
so the raw score of x = 64 corresponds to the standard z-score of z = -2
z = -2 means that the raw score x = 64 is 2 standard deviations below the mean (mu = 76)
now use a table to compute P(z < -2)
a table like this is usually found in the back of your stats book
Not the answer you are looking for? Search for more explanations.
you can also use the `normalcdf` function found on TI83, TI84, etc calculators
but your teacher will probably want you to use the table instead
How about if you are finding p(x is GREATER THAN or equal to 70)
-same mean and standard deviation :)
P(x > k) = 1 - P(x < k)
where k is some fixed number
using the rule, we know that
P(x > 70) = 1 - P(x < 70)
so you need to follow the steps to compute P(x < 70) first
then subtract that result from 1
oh and it doesn't matter if you have `greater than` or `greater than or equal to` since the two lead to the same answer
Okay last question: what if it has
p(x is greater than or equal to 85 but less than or equal to 91.
-I got .93 and .43 as the "Z's" for both of them, then I tried to subtract them from each other: .93-.43, but I think I did it wrong. (Sorry if that doesn't make any sense and also thank you so much for your help).
\[\Large P\left(a < x < b\right) = P(x < b) - P(x < a)\]
in this case, `b > a`