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4. Assume that a randomly selected subject is given a score. Those scores are normally distributed...

4. Assume that a randomly selected subject is given a score. Those scores are normally distributed with mean 0 and standard deviation 1. In each case, draw the graph (optional), then find the probability of the given scores. ROUND YOUR ANSWERS TO 4 DECIMAL PLACES

a. Find the probability of selecting a subject whose score is less than 1.13. ____________

b. Find the probability of selecting asubject whose score is greater than -1.28. ___________

c. Find the probability of selecting a subject whose score is between -0.54 & 2.07. _______

d. Find the probability of selecting a subject whose score is less than 0. ____________

e.Find the probability of selecting a subject whose score is greater than 0.

math   Statistics-And-Probability   
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Answer #1

Solution:

Part a) Find the probability of selecting a subject whose score is less than 1.13.

P( Z < 1.13) =.............?

P(Z< 1.13) Mean z=1.13

Look in z table for z = 1.1 and 0.03 and find corresponding area.

11 .0 .01 .02 ,04 ,05 .06 .07 .08 .09k | 0.0 5000 .5040 .5080 520 .5160 5199 .5239 .5279 5319 .5359 | 0.1 .5398 5438 5478 591

P( Z< 1.13) = 0.8708

Part b)  Find the probability of selecting asubject whose score is greater than -1.28.

P( Z> -1.28) =............?

P( Z> -1.28) = 1 - P( Z< -1.28)

P(Z >-1.28) z=-1.28 Mean

Look in z table for z = -1.2 and 0.08 and find corresponding area.

| 10010 7 .00 .01 -3.4 .0003 ,0003 -3.3 .0005 .0005 -3.2 | 0007 .0007 -3.1 .0010 1 | 0009 -3.0 .0013 0013 -2.9 .0019 0018 -2.

P( Z< -1.28) = 0.1003

Thus

P( Z> -1.28) = 1 - P( Z< -1.28)

P( Z> -1.28) = 1 - 0.1003

P( Z> -1.28) = 0.8997

Part c) Find the probability of selecting a subject whose score is between -0.54 & 2.07.

P( -0.54 < Z< 2.07 ) =P( Z < 2.07) - P( Z < -0.54)

P(0.54 <Z<2.07) z=-0.54 Mean z=2.07

Look in z table for z = 2.0 and 0.07 as well as for  z = -0.5 and 0.04  and find corresponding area.
[i ,00 .01 .02 .03 .04 .05 .06 1.07 .08 .09 10.0 .5000 5040 5080 .5120 5160 .5199 .5239 79 15319 5359 10.1 5398 .5438 5478 .5

P( Z<2.07 ) = 0.9808

and

-2.8 .00 .01 .02 .03 .04 .05 .06 .07 .08. .09 3.4 .0003 ,0003 ,0003 .0003 .0003 .0003 .0003 .0003 .0003 | 0002 -3.3 .0005 ,00

P( Z< -0.54) = 0.2946

thus

P( -0.54 < Z< 2.07 ) =P( Z < 2.07) - P( Z < -0.54)

P( -0.54 < Z< 2.07 ) = 0.9808 - 0.2946

P( -0.54 < Z< 2.07 ) = 0.6862

Part d) Find the probability of selecting a subject whose score is less than 0.

P( Z< 0.00) = .............?

P(Z<0.00) Mean

Look in z table for z = 0.0 and 0.00 and find corresponding area.

.00 0.0 0.1 0.2 0.3 0.4 0.5 15000 5398 .5793 .6179 .6554 .6915 | .01 1,5040 .5438 ,5832 .6217 .6591 16950 .02 .5080 .5478 587

P( Z< 0.00 ) = 0.5000

Part e) .Find the probability of selecting a subject whose score is greater than 0.

P( Z > 0.00) =.............?

P( Z > 0.00) = 1 - P( Z< 0.00)

P( Z > 0.00) = 1 - 0.5000

P( Z > 0.00) = 0.5000

P(Z > 0.00) Mean

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