Rogcr3ogsla2k
Answered

Find a value of the standard normal random variable Z, Call it Z0, such that
a. P(Z ≤ Z0) = .0401
b. P(- Z0 ≤ Z ≤ Z0) = .95
c. P(- Z0 ≤ Z ≤ Z0) = .90
d. P(- Z0 ≤ Z ≤ 0) = .2967

The 0 is subzero, don't know why the site couldn't show it.

Answer :

MissPhiladelphia
For the answer to this questions,a. P (z ≤ z0) = 0.0401=P(z ≥ z0) = 1-0.0401 = 0.9599 = P(z ≤ -z0) = 0.9599 
From tables z0 = -1.75 

b. P (-z0 ≤ z ≤ z0) = .95 = P (z ≤ z0)- P (z ≤ -z0) = P (z ≤ z0)- P (z ≥ z0) = 
P (z ≤ z0)-(1- P (z ≤ z0)) 
P (z ≤ z0) = (0.95+1)/2=0.975 
From tables z0 = 1.96 

c. P (-z0 ≤ z ≤ z0) = 0.90 
the procedure is the same that exercise b P (z ≤ z0) = (0.9+1)/2=0.95 
From tables the nearest value is z0 = 1.64 

d. P (-z0 ≤ z ≤ 0) = 0.2967= P (z ≤ 0) - P (z ≤ -z0) = P (z ≤ 0) - P (z ≥ z0) = 
P (z ≤ 0) - (1- P (z ≤ z0)) 
P (z ≤ z0) = 0.2967 + 1 - P (z ≤ 0)= 0.2967 + 1 - 0.5 = 0.7967 
From tables z0 = 0.83 

I hope my answer helped you

Using the normal distribution, it is found that:

a) [tex]Z_0 = -1.75[/tex].

b) [tex]Z_0 = 1.96[/tex].

c) [tex]Z_0 = 1.645[/tex].

d) [tex]Z_0 = -0.83[/tex].

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

  • Each z-score has a p-value associated with it, which is the percentile of X, and is found at the z-table.

Item a:

This is Z with a p-value of 0.0401, thus [tex]Z_0 = -1.75[/tex].

Item b:

  • Due to the symmetry of the normal distribution, the middle 95% is between the 2.5th and the 97.5th percentile.
  • We want the positive value, so Z with a p-value of 0.975, which is [tex]Z_0 = 1.96[/tex]

Item c:

Same logic as b, just middle 90%, thus [tex]Z_0 = 1.645[/tex].

Item d:

This is Z with a p-value of 1 - 0.2967 = 0.2033, thus [tex]Z_0 = -0.83[/tex].

A similar problem is given at https://brainly.com/question/12982818

Other Questions