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3 years ago

A. For exercises 1 and 2, a set of data has a normal distribution with a mean of 120 and a standard deviation of 10.

1 answer:

4 0

A1.) the interval about the mean within which 90% of the data lie = 120 + or - 1.645(10) = 120 + or - 16.45 = 120 - 16.45 to 120 + 16.45 = 103.55 to 136.45

2.) P(100 < X < 140) = P(X < 140) - P(X < 100) = P(z < (140 - 120)/10) - P(z < (100 - 120)/10) = P(z < 2) - P(z < -2) = P(z < 2) - [1 - P(z < 2)] = 2P(z < 2) - 1 = 2(0.97725) - 1 = 1.9545 - 1 = 0.9545 = 95.5%

B1.) the interval about the sample mean that has a 1% level of confidence is 500 + or - 2.575(80/√1000) = 500 + or - 7 = 500 - 7 to 500 + 7 = 493 to 507

2.) 2P(z < a) - 1 = 0.90
2P(z < a) = 1.90
P(z < a) = 0.95
a = 1.645
(b - 500)/(80/√1000) = 1.645
b - 500 = 4
b = 500 + 4 = 504
The interval about the sample mean such that the probability is 0.90 that the mean number lies within the interval is 496 - 504.

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$\bf ~~~~~~\textit{parabola vertex form}
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\begin{array}{llll}
\boxed{y=a(x- h)^2+ k}\\\\
x=a(y- k)^2+ h
\end{array}
\qquad\qquad
vertex~~(\stackrel{}{ h},\stackrel{}{ k})\\\\
-------------------------------\\\\
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3 years ago

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wlad13 [49]

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Step-by-step explanation:

Hey There!

So they want us to find the sine of angle S

well remember is sohcahtoa

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