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13 June, 14:59

Given that z is a standard normal random variable, find z for each situation.

a. The area to the left of z is. 9750

b. The area between 0 and z is. 4750

c. The area to the left of z is. 7291

d. The area to the right of z is. 1314

e. The area to the left of z is. 6700

f. The area to the right of z is. 3300 (Hint: Normal probability dist)

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  1. 13 June, 15:12
    0
    You have to use your standard normal table (or calculator) for these. The area under the curve should be 1, a z-value corresponds to the probability and is area to the left of that z.

    a. A simple reverse lookup of 0.9750. z=1.96

    b. Total area up to z would be 0.5+0.4750, so reverse look up 0.9750 and find z=1.96

    c. A simple reverse lookup. z=0.61

    d. Area to the left of z would b 1-0.1314, so lookup 0.8686 and find z=1.12

    e. A simple reverse lookup. z=0.44

    f. Just invert the area: 1-0.33 = 0.67 and reverse lookup: z=0.44
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