+A Read aloudFit to pagePage view|IX$666.10$593.84$521.580.920.0420.0420.46- Z-1.751.75archhpfafsf6T8hofoASt44$4&357YP008LE +Fit to pagence interval for a mean. We use the formula for a mcontinuous random variable. The point estimate forthe true population standard deviation because with 8the confidence interval.xt(a/2)Vnto the formula, we have:593.84 +1.75369.34V80иstandard normal table by looking up 0.46 in the body of tn the side and top of the table; 1.75. The solution for the in= 593.84 + 72.2636 = (521 .57, 666.10)u$ 521.58< u$ 666.108

Question
Asked Nov 14, 2019
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How did they get the 1.75? Using Confidence intervals. Please explain don’t understand how they used the z table they said they used .46

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$666.10
$593.84
$521.58
0.92
0.04
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0.46
- Z
-1.75
1.75
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+ A Read aloud Fit to page Page view | I X $666.10 $593.84 $521.58 0.92 0.04 2 0.04 2 0.46 - Z -1.75 1.75 arch hp fa fs f6 T8 ho fo ASt 44 $ 4 & 3 5 7 Y P 00 8 LE

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+
Fit to page
nce interval for a mean. We use the formula for a m
continuous random variable. The point estimate for
the true population standard deviation because with 8
the confidence interval.
xt(a/2)
Vn
to the formula, we have:
593.84 +1.75369.34
V80
и
standard normal table by looking up 0.46 in the body of t
n the side and top of the table; 1.75. The solution for the in
= 593.84 + 72.2636 = (521 .57, 666.10)
u
$ 521.58< u
$ 666.10
8
help_outline

Image Transcriptionclose

+ Fit to page nce interval for a mean. We use the formula for a m continuous random variable. The point estimate for the true population standard deviation because with 8 the confidence interval. xt(a/2) Vn to the formula, we have: 593.84 +1.75369.34 V80 и standard normal table by looking up 0.46 in the body of t n the side and top of the table; 1.75. The solution for the in = 593.84 + 72.2636 = (521 .57, 666.10) u $ 521.58< u $ 666.10 8

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Expert Answer

Step 1

For 92% confiden...

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(1a)100% 92% a 1-0.92 0.08 a 0.08 2 2 =0.04

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