what is p(z)=0.88

Q: Find z given that P(.24 <Z <z) = .12

A:

Q: If P(- b &lt; z &lt; b) = 0.4726, find b. b=

A:

Q: p(z < 1.29)

A: standard normal distribution Z~N(0,1) P(z&lt;1.29) = ?

Q: Calculate P (-0.1 < Z < 0.1). (Round to the hundredths).

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: X~N(1.8, 0.4).  Find P(0.9 < X < 2.3).  (Round to 3 decimal places.)

A:

Q: Calculate P (Z < 0). (Round to the hundredths).

A: GivenP(Z≤0)=

Q: what is M<E

A: Solution-As the solution is given,

Q: X 0 1 2 3 P(X = x) |0.512 0.384 0.096 0.008 Find P(more than 2 identifications are incorrect).

A: From the provided information, x 0 1 2 3 P (X = x) 0.512 0.384 0.096 0.008…

Q: P(Z<1.89) (less than or equal to)

A: Given that.   X~N( 0 , 1 )   μ=0  , ?=1 (for standard normal distribution)    Z-score =( x - μ )/?

Q: Find the value for z if P( Z > z ) = 0.7643.     0.63     -0.72     0.72     -0.63

A: Given that, P-value = 0.7643 Right tailed

Q: Calculate P(Z < 1.5)

A:

Q: Suppose P(E) = 0.284, P(F) = 0.222, and P(F ∩ E) = 0.129. What is P(F | E)? (Enter answer as a…

A: According to the given information in this question, we are going to find the value of P(F|E) by the…

Q: P(1.15 < Z < 2.77) =

A: X~N( 0 , 1 )   μ=0 , ?=1 (for standard normal distribution)   Z-score =( x - μ )/?

Q: P(z < -2.61)

A: Z score = -2.61 Left tailed test

Q: P (Z<1.8572)

A: Answer - Find   P (Z&lt;1.8572)

Q: P(0.51 ≤ Z ≤ 1.4) =

A: Draw the normal curve and shade the area between z=0.51 and z=1.4.

Q: P(X) = 0.62, P(Y) = 0.28, P(X and Y) = 0.16. Find P(not YIX). O none of the above 0.4129 0.5714…

A:

Q: If Z~N(0,1), then what is Pr(-0.09 < Z < 2.54)? You can use Excel. Enter your answer as a decimal…

A: We have to find given probability

Q: (h) P(1.26 ≤ Z ≤ 2.50) (1) P(1.90 ≤ Z) (1) P(|Z| ≤ 2.50)

A: It is given that the standard normal random variable Z with mean 0 and variance 1.   Note: The…

Q: The result for “Smoking makes me look older? Was F(1,517) = 17.02, p <0.001. Using this result,…

A: It is given that F(1,517) = 17.02. That is, the degrees of freedom for between sum of square is 1…

Q: P(−0.71 ≤ Z ≤ 1.34) =

A: Draw the normal curve and shade the area between z=-0.71 and z=1.34.

Q: P(X ≤ 1). n = 5, p = 0.6

A:

Q: p(x<=1) n=6 p=0.6

A: From the provided information, Sample size (n) = 6 Probability (p) = 0.6

Q: P(0.74<Z<2.11)

A: Let Z be the standard normal variable with parameters μ=0  and σ=1 The probability is,

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: suppose that X~N (2, oʻ), AP (2<X<4) =0.3, then P(X<0)= A.0.3 B.0.2 C.0.8 D.0.1

A:

Q: participants are there? 69 70 71 There is not enough information to determine given

A: t(70)=1.70 Degree of freedom =df =70