X ~ N (1, 4). Find the value of a since P(-2.5

Q: < Q = 12 Find the measure of < JLP. 4.

A:

Q: Find P(-1.40 < Z < 0)     .4192     .1259     .4357     .8907

A: Solution: Let Z be the standard normal random variable with mean μ=0 and standard deviation σ=1…

Q: 4 Compute f(g(x)) where f(x) =x* -x and g(x) =x²-11. dx The answer is

A:

Q: Calculate P(z<-1.2)

A: The z-score is standard normal variable with mean 0 and standard deviation 1.

Q: Enter the value of L- 1.27].

A: Floor value of x  is given as the greatest integer that is less than or equal to x

Q: Find P (-0. 55 < z < 2.51)

A:

Q: P(X ≤ 1). n = 5, p = 0.6

A:

Q: P(-1.45 < Z < +1.45) =?

A: Draw the normal curve that represents between z=-1.45 &amp; z=+1.45.

Q: 6. Which is the best list of possible values of x that could make f(x) : 3x + 6x? + 11x+ 6 equal to…

A: Solve the problem by substituting values in the problem. See if the value are larger than 100 then…

Q: P(- c ≤ Z ≤ c)=0.9399

A: We have given that  P(- c ≤ Z ≤ c)=0.9399 From the given information we have to find the value of c

Q: n=8,p=0.2,x<4

A:

Q: Let P(A) = 0.2 and P(B) = 0.5. If P(A|B) = 0.3, calculate P(A and B

A: Given data, P(A) = 0.2   P(B) = 0.5 P(A|B) = 0.3 P(A and B)=?

Q: P(-2.93&lt;z&lt;0)= P(-2.53&lt;z&lt;-0.240)=

A: The formula to calculate probability is,

Q: W. b) Let X₁, X₂X128G(2,4), by CLT find P(7 < X < 9) use (N(2)-0.977)

A:

Q: Find: x=  m<P= m<Q=

A: We have to solve

Q: P(-.80&lt;z&lt;.80)

A: Solution: The probability value is

Q: Suppose the cumulative distribution function of the random variable X is 0 x < 0 F(x) - 0 < x < 5 1…

A:

Limits

When it comes to calculus, limits are considered to be a very important topic of discussion. The main importance of limit lies in expressing the value of a function as it gets closer to a certain value. Also, limits can help you to understand other topics, such as continuity and differentiability better. In a broader sense, every calculus notion is a limit. Other branches of calculus have also been derived from limits.