You want to find
P(1000 < X < 3000)
where X is normally distributed with mean 1751 and standard deviation 421. Transform X to Z, so that it follows the standard normal distribution with mean 0 and standard deviation 1 using the relation
X = 1751 + 421Z ==> Z = (X - 1751)/421
Then
P(1000 < X < 3000) = P((1000 - 1751)/421 < (X - 1751)/421 < (3000 - 1751)/421)
… ≈ P(-1.783 < Z < 2.967)
… ≈ P(Z < 2.967) - P(Z < -1.783)
… ≈ 0.9985 - 0.0373
… ≈ 0.9612
so that approximately 96.1% of the students fall in this income range.
Answer:
yes, it is x=1/5
Step-by-step explanation:
i can give you the steps if you want
Answer: If you are solving for m it would be -37 or 37 with the added negative
Step-by-step explanation:
34 = -(m + 3)
34 = -m - 3
+ 3 = -m + 3
37 = -m
Answer:
23
Step-by-step explanation:
3 times 4 equals 12
then do
35 minus 12
Answer:
13x+38
Step-by-step explanation:
hope this'll help
