+ 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 + 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

Question

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
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-1.75
1.75
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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
View transcribed image text
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Transcribed Image Text

+ 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

+ 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