A. P(5.4 < x < 12.3)- B. P(x = 7.2) = 1

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

A:

Q: Calculate P ( Z > -0.1)

A:

Q: Prove that = Aatilag (Elog Xi) An. %3D

A:

Q: Assume P(Z>b)=0.22 for some number b. Without finding the value of b, what is P(Z<-b)?

A: From the provided information,P (Z&gt;b) = 0.22

Q: Find P(Z < 2.84)     .5375     .9575     .8098     .9977

A:

Q: P(z<2.25)=

A: Given Information: To find out Pz&lt;2.25

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: To find P(x<2.35) where do you shade?

A: Left side

Q: P(1.10 s z) P(IZ| s 2.50)

A: Here use normal distribution table for probability from z score

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: 0.13+15.04(x+3)=0.14+15.54

A: x=-1.96 Explanation is given below

Q: find P (Z > 1.21) = find P (Z < 1.60) = find P (0.59 < Z < 1.78) =

A: 1) P (Z &gt; 1.21) =?

Q: Assume P(Z> b) = 0.35 for some number b. Without finding the value of b, what P(Z < – b)?

A:

Q: find P(-1.2<Z<1.9)

A:

Q: P(x < 17.872)

A: Let's take the given  Sample size n=33 from a population, with mean=16 and standard deviation=8.4…

Q: Find P(X < 0.3)

A:

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: Calculate P(z<-1.2)

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

Q: P (Z<1.8572)

A: Answer - Find   P (Z&lt;1.8572)

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 ( -2.53 &lt; z &lt; 0.65)

A: The probability is,

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

A:

Q: p(0.7 <z < 1.31) =

A: Answer - Find P( 0.7 &lt; Z &lt; 1.31)  = ?

Q: what is p(z<?)=0.88

A: