Question

In: Statistics and Probability

Suppose Z denotes the standard normal random variable. Compute the following a) P (Z > 2)...

Suppose Z denotes the standard normal random variable. Compute the following

a) P (Z > 2)

b) P(-1 < Z < 2.31)

c) P (1.21 < Z < 2.42)

d) P (Z < 1.37)

e) P (Z > -1)

Show working

Solutions

Expert Solution

We can compute the probabilities using any software or standard normal tables. We would be using R here as:

a) P(Z > 2) = 1 - P(Z < 2)


Therefore 0.02275013 is the required probability here.

b) The probability here is computed as:
P( -1 < Z < 2.31) = P(Z < 2.31) - P(Z < -1)


Therefore 0.8309007 is the required probability here.

c) The probability here is computed as:
P( 1.21 < Z < 2.42) = P(Z < 2.42) - P(Z < 1.21)


Therefore 0.1053792 is the required probability here.

d) The probability here is computed as:
P(Z < 1.37)


Therefore 0.9146565 is the required probability here

e) The probability here is computed as:
P( Z > -1) = 1 - P(Z < -1)


Therefore 0.8413447 is the required probability here.


Related Solutions

Given that z is a standard normal random variable, compute the following probabilities. P(z ≤ -0.71)...
Given that z is a standard normal random variable, compute the following probabilities. P(z ≤ -0.71) P(z ≤ 1.82) P(z ≥ -0.71) P(z ≥ 1.22) P( –1.71 ≤ z ≤ 2.88) P( 0.56 ≤ z ≤ 1.07) P( –1.65 ≤ z ≤ –1.65) Given that z is a standard normal random variable, find z, for each situation. The area to the left of z is 0.9608 The area to the right of z is .0102 The area between o and...
If P(-2 < Z < k)=.6 , where Z is a standard normal random variable, then...
If P(-2 < Z < k)=.6 , where Z is a standard normal random variable, then k is... Select one: a. 0.195 b. 0.73 c. 0.55 d. -0.40
Z is a standard normal random variable, then k is ... a. P(Z < k) =...
Z is a standard normal random variable, then k is ... a. P(Z < k) = 0.92 b. P(Z > k) = 0.72 c. P(Z ≤ k) = 0.26 d. (A-Grade) P(−1 < Z < k) = 0.60 e. (A-Grade) P(k < Z < 1.7) = 0.57 f. (A-Grade) P(Z = k) = 0.00
Find the following probabilities for the standard normal random variable z z : a) P(−2.07≤z≤1.93)= P...
Find the following probabilities for the standard normal random variable z z : a) P(−2.07≤z≤1.93)= P ( − 2.07 ≤ z ≤ 1.93 ) = (b) P(−0.46≤z≤1.73)= P ( − 0.46 ≤ z ≤ 1.73 ) = (c) P(z≤1.44)= P ( z ≤ 1.44 ) = (d) P(z>−1.57)= P ( z > − 1.57 ) =
3. A) Given that z is a standard normal random variable, compute the probability that it...
3. A) Given that z is a standard normal random variable, compute the probability that it takes on a value between -2 and -1. 3. B). Given that z is a standard normal random variable, find the z-score for a situation where the area to the right of z is 0.0901.
Given that z is a standard normal random variable, compute the probability that it takes on...
Given that z is a standard normal random variable, compute the probability that it takes on a value between -2 and -1.
Find the following probability for a standard normal random variable, P(Z ≥ -2.16 )
Find the following probability for a standard normal random variable, P(Z ≥ -2.16 )
Let z be a random variable with a standard normal distribution. Find “a” such that P(|Z|...
Let z be a random variable with a standard normal distribution. Find “a” such that P(|Z| <A)= 0.95 This is what I have: P(-A<Z<A) = 0.95 -A = -1.96 How do I use the symmetric property of normal distribution to make A = 1.96? My answer at the moment is P(|z|< (-1.96) = 0.95
Let z be a random variable with a standard normal distribution. Find P(0 ≤ z ≤...
Let z be a random variable with a standard normal distribution. Find P(0 ≤ z ≤ 0.46), and shade the corresponding area under the standard normal curve. (Use 4 decimal places.)
1. If Z is a standard normal random variable, find c such that P(−c ≤ Z...
1. If Z is a standard normal random variable, find c such that P(−c ≤ Z ≤ c) = 0.82. [Answer to 2 decimal places] 2. Weakly earnings on a certain import venture are approximately normally distributed with a known mean of $353 and unknown standard deviation. If the proportion of earnings over $386 is 25%, find the standard deviation. Answer only up to two digits after decimal. 3. X is a normal random variable with mean μ and standard...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT