Solutions to Homework #6
(W4150 ‘Intro to Probability and Statistics’, S04)
Sec. 6.4. (1)
Use tables or builtin Excel function for normal distribution to answer each question here:
(a) Area =
Φ
(1.43) = .9236
(b) Area = 1 
Φ
(.89) = .8133
(c) Area =
Φ
(.65) 
Φ
(2.16) = .2424
(d) Area =
Φ
(1.39) = .0823
(e) Area = 1 
Φ
(1.96) = .0250
(f) Area =
Φ
(1.74) 
Φ
(.48) = .6435
Sec.6.4. (3)
(a) From table A.3, k = 1.72
(b) Since P(Z > k) = .2946 then P(Z < k) = 1 – P(Z > k) = .7054; and from the same table we
find k = .54
(c) The area to the left of z = .83 is found from table A.3 (or Excel) to be .1762. Therefore,
the total area to the left of k is (.1762) + (.7235) = .8997
k = 1.28
Sec. 6.4. (4)
(a) z = (17 – 30)/6 = 2.17
Area = 1 
Φ
(2.17) = 1  .0150 = .9850
(b) z = (22 – 30)/6 = 1.33
Area =
Φ
(1.33) = .0918
(c) z
1
= (32 – 30)/6 = .33, z
2
= (41 – 30)/6 = 1.83
Area =
Φ
(1.83) 
Φ
(.33) = .9664  .6293
= .3371
(d) z = .84
x = 6(.84) + 30 = 35.4
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 Spring '04
 GALLEGO
 Normal Distribution, Sec.

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