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
(e) z
1
= 1.15 and z
2
= 1.15
°
x
1
= 6(1.15) + 30 = 23.1 and x
2
= 6(1.15) + 30 = 36.9
Sec. 6.4. (6)
Z
1
= [(
μ
 3
σ
) 
μ
]/
σ
= 3; Z
2
= [(
μ
+ 3
σ
) 
μ
]/
σ
= 3.
Hence P(
μ
 3
σ
< X <
μ
+ 3
σ
) = P(3 < Z < 3) =
Φ
(3) 
Φ
(3) = .9987  .0013 = .9974
Sec. 6.4. (14)
Everywhere here you’re supposed to respect the fact that the heights were measured within
precision of +
.25cm (i.e. 0.5cm in total) and therefore every time you recalculate the r.v. Z
into standard normal you have to subtract .25cm from the actual value if you’re looking for
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 Spring '04
 GALLEGO
 Normal Distribution, Sec.

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