Question

The height of women ages 20-29 is normally distributed, with a mean of 64.6 inches. Assume o= 2.5 inches. Are you more likely
X i Standard Normal Table (Page 1) Area 09 .08 .07 .06 05 04 03 .02 .01 00 0003 0004 3.4 0002 0003 0003 0003 0004 0003 0004 0
Standard Normal Table (Page 1) -3.3 .0003 0004 0004 0004 0004 0004 0004 0005 0005 0007 0005 - 3.2 0005 0005 0006 0006 0006 00
i Standard Normal Table (Page 2) Area 0 .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 0.0 5000 5040 5080 5478 5120 5517 5160 5557 5
Standard Normal Table (Page 2) 0.1 5398 5438 5478 5517 5557 5596 5636 5675 5714 5753 5793 6179 0.2 5832 5871 5910 5948 5987 6
0 0
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Answer #1

Solution :

Given that ,

mean = \mu = 64.6

standard deviation = \sigma = 2.5

a) P(x < 66.2) = P[(x - \mu ) / \sigma < (66.2 - 64.6) / 2.5]

= P(z < 0.64)

Using z table,

= 0.7389

b) n = 21

\mu\bar x = \mu = 64.6

\sigma\bar x = \sigma / \sqrt n = 2.5 / \sqrt 21 = 0.546

P(\bar x < 66.2) = P((\bar x - \mu \bar x ) / \sigma \bar x < (66.2 - 64.6) / 0.546)

= P(z < 2.93)

Using z table

= 0.9983

c) correct option is = D.

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