b. P[2570 a]= 0.508 Z-score: 9.78 P[Za]= 0

Q: P(x > 29 | x < 51) =

A:

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

A: We have to find given probability.

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: 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: P(z>1.2)

A: Given problem Given that   P(z&gt;1.2)   Ans) P( z&gt;1.2 ) is, = 1 - P( z ≤ 1.2 )

Q: Find P(F<2.95) nu1=12 nu2=12 Make sure to show your work.

A: nu1=12nu2=12

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

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

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: Victims of Violence A random survey of 70 women who were victims of violence found that 28 were…

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(-1.33<z<0)= round answer to four decimal places as needed

A: The z-scores are -1.33 and 0.

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: (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: H0: 0.63  Ha:<0.63 a= 0.05 p-hat= 0.592 n= 250 whats the Z-score? whats the P-value?

A:

Q: p(0 < z < 1.30)= p(0 < z < 1.83)= p(1.30 < z < 1.83)=

A: Given that:Z is standard normal Distributed with i.e. mean µ =0 and standard deviation σ =1 By…

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(Z ≤ 0.77) = [] P(Z > 1.36) = [ P(−0.70 < Z < 2.06) = |

A: Given P(z≤0.77)

Q: p(0.7 <z < 1.31) =

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