Question 271767: Given a normal distribution with a mean of 100 and a standard deviation of 10 what is the probability that a) x>75 b)x<70 c)x>80 or x>110 d) 80% of the data are between what two values(symmetrically distributed around the mean)
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Given a normal distribution with a mean of 100 and a standard deviation of 10 what is the probability that
a) x>75
Use a TI calculator to get:
z(75) = (75-100)/10 = -25/10 = -2.5
P(x> 75) = P(z > -2.5) = normalcdf(-2.5,100) = 0.9938
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b)x<70
Samee procedure:
1st: Find the z-value for 70.
Find P(z < -3) = normalcdf(-100,-3) = 0.0013..
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c)x>80 or x>110
x> 80 inclues all the x> 110
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z for 80 is +1
P(z > 1) = normalcdf(1,100) = 0.1587
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d) 80% of the data are between what two values(symmetrically distributed around the mean)
Find the z-values that bracket 80% centered at the mean:
That area leaves tails of 10% on the right and on the left under
the normal curve.
Find invNorm(0.1) = -2.3263
So z = -2.3263 and z=+2.3263
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Now find the x-value corresponding to those z-values
x = zs + u
x = -2.3263*10 + 100 = 76.74
and
x = +2.3263*10 + 100 = 123.26
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Cheers,
Stan H.
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