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uranmaximum [27]
3 years ago
10

99 POINTS HELP ME HURRY

Mathematics
1 answer:
Stels [109]3 years ago
7 0

Answer:

c = sqrt(11)

Step-by-step explanation:

We can use the pythagorean theorem to calculate  the length of the hypotenuse.

c^2 = a^2 +b^2  where a and b are the legs and c is the hypotenuse

c^2 = 3^2 + (sqrt(2))^2

c^2 = 9 + 2

c^2 =11

Take the square root of each side.

sqrt(c^2) = sqrt(11)

c = sqrt(11)

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The weights of 83 randomly selected windshields were found to have a variance of 1.88. Construct the 95% confidence interval for
Morgarella [4.7K]

Answer:

95% confidence interval for the population variance = (1.42 , 2.62).

Step-by-step explanation:

We are given that the weights of 83 randomly selected windshields were found to have a variance of 1.88.

<em>So, firstly the pivotal quantity for 95% confidence interval for the population variance is given by;</em>

        P.Q. = \frac{(n-1)s^{2} }{\sigma^{2} } ~ \chi^{2} __n_-_1

where, s^{2} = sample variance = 1.88

           \sigma^{2} = population variance

            n = sample of windshields = 83

So, 95% confidence interval for population variance, \sigma^{2} is;

P(58.85 < \chi^{2} __8_2 < 108.9) = 0.95 {As the table of \chi^{2} at 82 degree of freedom

                                              gives critical values of 58.85 & 108.9}

P(58.85 < \frac{(n-1)s^{2} }{\sigma^{2} } < 108.9) = 0.95

P( \frac{ 58.85}{(n-1)s^{2} } < \frac{1 }{\sigma^{2} } < \frac{108.9}{(n-1)s^{2} } ) = 0.95

P( \frac{ (n-1)s^{2}}{108.9 } < \sigma^{2} < \frac{ (n-1)s^{2}}{58.85 } ) = 0.95

<em><u>95% confidence interval for</u></em> \sigma^{2} = ( \frac{ (n-1)s^{2}}{108.9 } , \frac{ (n-1)s^{2}}{58.85 } )

                                                  = ( \frac{ (83-1)\times 1.88}{108.9 } , \frac{ (83-1)\times 1.88}{58.85 } )

                                                  = (1.42 , 2.62)

Therefore, 95% confidence interval for the population variance of the weights of all windshields in this factory is (1.42 , 2.62).

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