Let X ~ Normal ( =3 , 2 =16) and Z ~ Normal (0,1), use the Standard Normal table
ID: 3153720 • Letter: L
Question
Let X ~ Normal (=3,2=16) and Z ~ Normal (0,1), use the Standard Normal table to compute the following:1. P(|X| > 2) 2. P(0.5 < X< 1) 3. P(-2 < X < +2) 4. The point z such that P(Z z)=0.8 Let X ~ Normal (=3,2=16) and Z ~ Normal (0,1), use the Standard Normal table to compute the following:
1. P(|X| > 2) 2. P(0.5 < X< 1) 3. P(-2 < X < +2) 4. The point z such that P(Z z)=0.8 1. P(|X| > 2) 2. P(0.5 < X< 1) 3. P(-2 < X < +2) 4. The point z such that P(Z z)=0.8
Explanation / Answer
Given that X ~ Normal (=3,2=16) and Z ~ Normal (0,1), use the Standard Normal table to compute the following:
1) P(|x| > 2) :
P(|x| >2) = P(-2 < x < 2)
convert x=-2 and x=2 into z-score.
z = (x - mean) /sd
z for x = -2 :
z = (-2 - 3)/4 = -1.25
z for x=2 :
z = (2-3)/4 = -0.25
That is now we have to find P(-1.25 < Z < -0.25).
P(-1.25 < Z < -0.25) = P(Z <= -0.25) - P(Z <= -1.25)
This probability we can find by using EXCEL.
syntax is,
=NORMSDIST(z)
where Z is test statistic value.
P(-1.25 < Z < -0.25) = 0.4013 - 0.1056 = 0.2956
2. P(0.5 < X< 1) :
z-score for x=0.5 :
z = (0.5 - 3) / 4 = -0.625
z-score for x=1 :
z = (1 - 3) / 4 = -0.5
That is now we have to find P(-0.625 < Z < -0.5).
P(-0.5 < Z < -0.375) = P(Z <=-0.5) - P(Z <= -0.625)
= 0.3085 - 0.2660
= 0.0426
3. P(-2 < X < +2) :
z-score for x = -2 is,
z = ( (-2 - ) / ) = -2
z-score for x = +2 is,
z = ( (+2 - ) / ) = 2
That is we have to find P(-2 < Z < 2).
P(-2 < Z < 2) = P(Z<=2) - P(Z<=-2)
= 0.9772 - 0.0228
= 0.9545
4. The point z such that P(Z z) = 0.8
1 - P(Z <= z) = 0.8
P(Z<=z) = 0.2
Now we have to find z when probability is 0.2.
This we can find by using EXCEL.
syntax is,
=NORMSINV(probability)
where probability = 0.2
z = -0.84
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