Question
Asked Oct 20, 2019

Find the percentage of data points that lie between -2.31 and 1.85.

check_circleExpert Solution
Step 1

Assuming the data follows standard normal distribution with mean 0 and standard Deviation 1
Finding the p-value at z = -2.31 = 0.010444 (From Excel = NORM.S.DIST(-2.31,TRUE)

L0
-2.31
Р(X <х) %3
0.01044
0.4
0.3
0.2
0.1
0.0
-4
2
2
-3
-1
0
1
X
2
o SD(X) 1
Var(X) 1
E(X) 0
4
(x)
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L0 -2.31 Р(X <х) %3 0.01044 0.4 0.3 0.2 0.1 0.0 -4 2 2 -3 -1 0 1 X 2 o SD(X) 1 Var(X) 1 E(X) 0 4 (x)

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Step 2

Finding the p-value at z = 1.85 = 0.967843 (From Excel = NORM.S.DIST(1.85,TRUE)

1
0
0.96784
1.85
Р(X <х) -
0.4
0.3
0.2
0.1
0.0
4
2
3
-1
1
2
2Var(X) 1
= E(X) = 0
a SD(X) 1
OX
(x)
help_outline

Image Transcriptionclose

1 0 0.96784 1.85 Р(X <х) - 0.4 0.3 0.2 0.1 0.0 4 2 3 -1 1 2 2Var(X) 1 = E(X) = 0 a SD(X) 1 OX (x)

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Step 3

Assuming the data follows standard normal distribution with mean 0 and standard Deviation 1
Finding the p-value at z = -2.31 = 0.010444 (From Excel = NORM.S.DIST(-2.31,TR...

= (Pz=1.85
Pz-2.31) x 100
= (0.967843 - 0.010444) x 100
95.7399 %
help_outline

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= (Pz=1.85 Pz-2.31) x 100 = (0.967843 - 0.010444) x 100 95.7399 %

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