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Virty [35]
3 years ago
9

Ans this question......​

Mathematics
1 answer:
Helga [31]3 years ago
6 0

Step-by-step explanation:

Universal Set(U)={2,3,4...16}

M={2,4,6,8..16}[Even numbers between 2-16]

N={3,5,7...15}[Odd numbers between 2-16]

(PUM)'=U-(PUM)={2,3,4...16}-[{2,4,6,8..16}U{3,5,7...15}]={}

For shading, Shade the area outside the two circles but inside the square.

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The heights of 40 randomly chosen men are measured and found to follow a normal distribution. An average height of 175 cm is obt
AVprozaik [17]

Answer:

95% two-sided confidence interval for the true mean heights of men is [168.8 cm , 181.2 cm].

Step-by-step explanation:

We are given that the heights of 40 randomly chosen men are measured and found to follow a normal distribution.

An average height of 175 cm is obtained. The standard deviation of men's heights is 20 cm.

Firstly, the pivotal quantity for 95% confidence interval for the true mean is given by;

                             P.Q. = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample average height = 175 cm

            \sigma = population standard deviation = 20 cm

            n = sample of men = 40

<em>Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.</em>

So, 95% confidence interval for the true mean, \mu is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5%

                                     level of significance are -1.96 & 1.96}  

P(-1.96 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 1.96) = 0.95

P( -1.96 \times }{\frac{\sigma}{\sqrt{n} } } < {\bar X-\mu} < 1.96 \times }{\frac{\sigma}{\sqrt{n} } } ) = 0.95

P( \bar X-1.96 \times }{\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+1.96 \times }{\frac{\sigma}{\sqrt{n} } } ) = 0.95

<u>95% confidence interval for </u>\mu = [ \bar X-1.96 \times }{\frac{\sigma}{\sqrt{n} } } , \bar X+1.96 \times }{\frac{\sigma}{\sqrt{n} } } ]

                                            = [ 175-1.96 \times }{\frac{20}{\sqrt{40} } } , 175+1.96 \times }{\frac{20}{\sqrt{40} } } ]

                                            = [168.8 cm , 181.2 cm]

Therefore, 95% confidence interval for the true mean height of men is [168.8 cm , 181.2 cm].

<em>The interpretation of the above interval is that we are 95% confident that the true mean height of men will be between 168.8 cm and 181.2 cm.</em>

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