(14) Use a table of the distribution of N(0, 1) to find P(-1.28 < X < 1.96). The correct answer is 0.975 0.87 0.025 -0.025 -0.87 N/A (Select One) 오이

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section: Chapter Questions
Problem 52RE
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(13) Let X have a chi square density with n = 4 degrees of freedom. Show that its cdf is:
2–1
(t/2)
F() 3D 1 — еxp(-1/2) >
j!
j=0
t > 0.
Here is the proof. For any real number t > 0,
F(t) =
T(2)
(1/2)?
x2-1 e-x2 dx,
–x/2
-2te¬/2
+2
dx
(ii)
(-2te-2 + 4(1 – e42))
= 1 - e 12 – (1/2)e¬2.
Choose the best justification of the two steps (i) and (ii).
Newton's law
Triangle inequality
Partial fractions
Integration by parts
Definition of cdf
N/A
(Step (i) because of)
Newton's law Triangle inequality Partial fraction Integration by parts Definition of cdf N/A
(Step (ii) because of) O
(14) Use a table of the distribution of N(0, 1) to find P(-1.28 < X < 1.96).
The correct answer is
0.975
0.87
0.025
-0.025
-0.87
N/A
(Select One)
Transcribed Image Text:(13) Let X have a chi square density with n = 4 degrees of freedom. Show that its cdf is: 2–1 (t/2) F() 3D 1 — еxp(-1/2) > j! j=0 t > 0. Here is the proof. For any real number t > 0, F(t) = T(2) (1/2)? x2-1 e-x2 dx, –x/2 -2te¬/2 +2 dx (ii) (-2te-2 + 4(1 – e42)) = 1 - e 12 – (1/2)e¬2. Choose the best justification of the two steps (i) and (ii). Newton's law Triangle inequality Partial fractions Integration by parts Definition of cdf N/A (Step (i) because of) Newton's law Triangle inequality Partial fraction Integration by parts Definition of cdf N/A (Step (ii) because of) O (14) Use a table of the distribution of N(0, 1) to find P(-1.28 < X < 1.96). The correct answer is 0.975 0.87 0.025 -0.025 -0.87 N/A (Select One)
Expert Solution
Step 1

14) Given Information:

Use a table of the distribution of N0,1 to find P-1.28<X<1.96

The z-score is normally distributed with a mean 0 and a standard deviation 1. It is known as the standard normal curve.

Z-score indicates how many standard deviations away, a data value is, from the mean. It is given by the formula:

Z=X-μσ

Z-score for X = -1.28 is:

Z=-1.28-01=-1.28

Z-score for X = 1.96 is 1.96

 

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