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Crank
3 years ago
7

A Statistics class is estimating the mean height of all female students at their college. They collect a random sample of 36 fem

ale students and measure their heights. The mean of the sample is 65.3 inches. The standard deviation is 5.2 inches. Use the T-distribution Inverse Calculator applet to answer the following question. What is the 90% confidence interval for the mean height of all female students in their school? Group of answer choices (56.5, 74.1) (63.6, 67.0) (63.8, 66.8) (63.9, 66.7)
Mathematics
1 answer:
Butoxors [25]3 years ago
4 0

Answer: = ( 63.9, 66.7)

Therefore at 90% confidence interval (a,b)= ( 63.9, 66.7)

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean x = 65.3

Standard deviation r = 5.2

Number of samples n = 36

Confidence interval = 90%

z(at 90% confidence) = 1.645

Substituting the values we have;

65.3 +/-1.645(5.2/√36)

65.3 +/-1.645(0.86667)

65.3+/- 1.4257

65.3+/- 1.4

= ( 63.9, 66.7)

Therefore at 90% confidence interval (a,b)= ( 63.9, 66.7)

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HELP PLEASE ASAP
Pie

\bold{\huge{\underline{ Solution }}}

<u>We </u><u>have</u><u>, </u>

  • Line segment AB
  • The coordinates of the midpoint of line segment AB is ( -8 , 8 )
  • Coordinates of one of the end point of the line segment is (-2,20)

Let the coordinates of the end point of the line segment AB be ( x1 , y1 ) and (x2 , y2)

<u>Also</u><u>, </u>

Let the coordinates of midpoint of the line segment AB be ( x, y)

<u>We </u><u>know </u><u>that</u><u>, </u>

For finding the midpoints of line segment we use formula :-

\bold{\purple{ M( x,  y) = }}{\bold{\purple{\dfrac{(x1 +x2)}{2}}}}{\bold{\purple{,}}}{\bold{\purple{\dfrac{(y1 + y2)}{2}}}}

<u>According </u><u>to </u><u>the </u><u>question</u><u>, </u>

  • The coordinates of midpoint and one of the end point of line segment AB are ( -8,8) and (-2,-20) .

<u>For </u><u>x </u><u>coordinates </u><u>:</u><u>-</u>

\sf{  -8  = }{\sf{\dfrac{(- 2 +x2)}{2}}}

\sf{2}{\sf{\times{ -8  = - 2 + x2 }}}

\sf{ - 16 = - 2 + x2 }

\sf{ x2 = -16 + 2 }

\bold{ x2 = -14  }

<h3><u>Now</u><u>, </u></h3>

<u>For </u><u>y </u><u>coordinates </u><u>:</u><u>-</u>

\sf{  8  = }{\sf{\dfrac{(- 20 +x2)}{2}}}

\sf{2}{\sf{\times{ 8   = - 20 + x2 }}}

\sf{ 16 = - 20 + x2 }

\sf{ y2 = 16 + 20 }

\bold{ y2 = 36  }

Thus, The coordinates of another end points of line segment AB is ( -14 , 36)

Hence, Option A is correct answer

7 0
2 years ago
#5-12 you don't need to answer just plz explain how to do it
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