Question

In: Statistics and Probability

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 ) =

Expert Solution

Solution:

Given that,

Using standard normal table;

(a)

P(-2.07 z 1.93)

= P(z 1.93) - P(z -2.07)

= 0.9732 - 0.0192

= 0.954

P(-2.07 z 1.93) = 0.954

(b)

P(-0.46 z 1.73)

= P(z 1.73) - P(z -0.46)

= 0.9582 - 0.3228

= 0.6354

P(-0.46 z 1.73) = 0.6354

(c)

P(z 1.44) = 1 - P(z 1.44) = 1 - 0.9251 = 0.0749

(d)

P(z > -1.57) = 1 - P(z < -1.57) = 1 - 0.0582 = 0.9418

orchestra answered 2 years ago

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 Z is a standard normal random variable, find the value z0 for the following probabilities....

If Z is a standard normal random variable, find the value z0 for the following probabilities. (Round your answers to two decimal places.) (a) P(Z > z0) = 0.5 z0 = (b) P(Z < z0) = 0.9279 z0 = (c) P(−z0 < Z < z0) = 0.90 z0 = (d) P(−z0 < Z < z0) = 0.99 z0 =

Find the following probabilities for the standard normal random variable z: (a) P(−0.76<z<0.75)= (b) P(−0.98<z<1.36)= (c)...

Find the following probabilities for the standard normal random variable z: (a) P(−0.76<z<0.75)= (b) P(−0.98<z<1.36)= (c) P(z<1.94)= (d) P(z>−1.2)= 2. Suppose the scores of students on an exam are Normally distributed with a mean of 480 and a standard deviation of 59. Then approximately 99.7% of the exam scores lie between the numbers ---- and -----. ?? Hint: You do not need to use table E for this problem.

Find the probabilities for the standard normal random variable z: (1 point) P(-2.58<z<2.58) x is a...

Find the probabilities for the standard normal random variable z: (1 point) P(-2.58<z<2.58) x is a normal random variable with mean (μ) of 10 and standard deviation (σ) of 2. Find the following probabilities: (4 points) P(x>13.5) (1 point) P(x<13.5) (1 point) P(9.4<x<10.6) (2 points)

Find the following probability for a standard normal random variable, P(Z ≥ -2.16 )

Find the following probabilities for a standard normal random variable Z. Note: Round your answers to...

Find the following probabilities for a standard normal random variable Z. Note: Round your answers to four decimal places. A) P(Z < -1.47) B) P(Z > 2.20) C) P(Z > -1.17) D) P(Z < 1.30)

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

) Find the following probabilities for the standard normal random variable zz: (a) P(−2≤z≤2.01)= (b) P(−0.5≤z≤1.62)= (c) P(z≤1.3)= (d) P(z>−1.1)=

1. Suppose a random variable, Z, follows a Standard Normal distribution. Find the following probabilities using...

1. Suppose a random variable, Z, follows a Standard Normal distribution. Find the following probabilities using the z-table. For the Z distribution, find the following. Use z-table and check with Excel. Sketch, etc. (a) the 28th percentile (b) the 59th percentile.

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.)