Solutions to Homework #7
(W4150 ‘Intro to Probability and Statistics’, S04)
Sec. 8.5. (1)
P(
μ
x
–1.9
σ
x
<
X <
μ
x
–.4
σ
x
) = P(
μ
x
–1.9
σ
x

μ
x
<
X 
μ
x
<
μ
x
–.4
σ
x

μ
x
) = P(1.9 < (X 
μ
x
)/
σ
x
< .4) = P(z
1
= 1.9 < Z < .4 = z
2
) = .3446  .0287 = .3159
Sec.8.5. (3)
(a) for n = 64,
σ
x
=
σ
/
√
n = 5.6/8 = .7, whereas for n = 196,
σ
x
= 5.6/14 = .4
the standard
error of the mean is reduced from .7 to .4 when the sample size is increased from 64 to
196
(b) for n = 784,
σ
x
=
σ
/
√
n = 5.6/28 = .2, whereas for n = 49,
σ
x
= 5.6/7 = .8
the standard
error of the mean is increased from .2 to .8 when the sample size is decreased from 784 to
49
Sec. 8.5. (6)
(a)
μ
x
=
μ
= 174.5,
σ
x
=
σ
/
√
n = 6.9/5 = 1.38 (given n = 25)
(b) z
1
= (172.45 – 174.5)/1.38 = 1.49; z
2
= (175.85 – 174.5)/1.38 = .98 where we
add/subtract .05 due to the given precision of measurement of 0.1 cm. Hence P(172.5
<
X < 175.9) = P(1.49 < Z < .98) = .8365  .0681 = .7684. And so the number of sample
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 Spring '04
 GALLEGO
 critical values, Sec., σx

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