If 2 < X < 4, X is between 2 and 4 but 2 & 4 is not included, which means X is more than 2 but not 2. Similarly X < 4 means X is less than 4 but not 4. So the only number that satisfy 2 < X < 4 is 3.
So, P(2<X<4) = P(X = 3) = 0.20
Let X N(1,3) and Y~ N(2,4), where X and Y are independent 1. P(X <4)-? P(Y < 1) =? 4、 5, P(Y < 6) =? 7, P(X + Y < 4) =?
2. The random variable X has density ke-lal. Find P(|X| + |X – 31 < 5). A. .716 B. .235 C. .807 D. .344
For #4-5, use the following information: Overhead reach distances X are used in planning assembly workstations. The overhead reach distance of adult females is assumed to be X ~ N (202.5 cm, 8.0 cm) where cm refers to centimeters. 4. If 1 adult female is randomly selected, find the probability that her overhead reach is between 194.5 cm and 210.5 cm. Use the graph to sketch the probability as the area under the pdf given. 5. If 64 adult females...
1. Let x, a € R. Prove that if a <a, then -a < x <a.
If X is a Poisson variable such that P(X=2)=3/10 and P(X=1)=1/5. Then P(0.2<X<2.9)+P(X=3.5) equal to A discrete random variable X has a cumulative distribution function defined by F(x) (x+k) for x = 0,1,2 Then the value of k is 16
Question 2 2 pt Find P(z<2.35)= Round to 4 decimal places.
Suppose that NoP(X5)6 and P(X 2):2 find P( 3< X < 4).
4. Find & such that |--^x=12,< for all \x + 2<8.
Question 4 2 pts Find P(-1.47< z< 1.79) = places. Round to 4 decimal
Apply Chebyshevs Inequality to lower bound P(O< X < 4) when E(X) 2 and E(X2)-5