Answer:
The probability that there are 3 or less errors in 100 pages is 0.648.
Step-by-step explanation:
In the information supplied in the question it is mentioned that the errors in a textbook follow a Poisson distribution.
For the given Poisson distribution the mean is p = 0.03 errors per page.
We have to find the probability that there are three or less errors in n = 100 pages.
Let us denote the number of errors in the book by the variable x.
Since there are on an average 0.03 errors per page we can say that
the expected value is,
= E(x)
= n × p
= 100 × 0.03
= 3
Therefore the we find the probability that there are 3 or less errors on the page as
P( X ≤ 3) = P(X = 0) + P(X = 1) + P(X=2) + P(X=3)
Using the formula for Poisson distribution for P(x = X ) = 
Therefore P( X ≤ 3) = 
= 0.05 + 0.15 + 0.224 + 0.224
= 0.648
The probability that there are 3 or less errors in 100 pages is 0.648.
Rhombus, Parallelogram, Kite, Rectangle, Square, Trapezoid, Isosceles Trapezoid
WidthAnswer:
Step-by-step explanation:
Answer:
$441
Step-by-step explanation:
Since each ticket costs 10.50 and you buy 42 :
10.5 * 42 = 441
Answer:
None
Step-by-step explanation:
There is no integer equivalent to 91/2, integers can't have a decimal.
However, I assume you would like me to answer 9.5, even though it isn't an integer.