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Answer :
a) The probability that her weight is between 113 lb and 176 lb is 0.7220.
b) The probability that they have a mean weight between 113 16 and 176 is 0.9391.
c) The probability that they have a mean weight between 113 ib and 176 lb is 1.0000.
From the given data,
Here,
µ = 143
σ = 29
a) P( 113 ≤ [tex]\bar x[/tex] ≤ 176 ) , n = 1
[tex]\mu\bar x[/tex] = 143
[tex]\sigma\bar x=\frac{\sigma}{\sqrt {n}}[/tex]
[tex]=\frac{29}{\sqrt{1}}[/tex]
[tex]P[\frac{( 113 - 143 )}{29} \le ( x - \frac {\mu\bar x} {\sigma\bar x})\le \frac{ ( 176 - 143 )} {29} ][/tex]
[tex]P( -1.0345 \le X \le 1.1379 )[/tex]
[tex]P( z < 1.1379 ) - P( Z < -1.0345 )[/tex]
[tex]0.8724 - 0.1505[/tex]
[tex]= 0.7220[/tex]
b) [tex]P( 113 \le \bar x \le 176 ) , n = 3[/tex]
[tex]= \mu \bar x[/tex]
= 143
[tex]= \sigma \bar x[/tex]
[tex]= \frac{\sigma}{\sqrt{n}}[/tex]
[tex]= \frac{29}{\sqrt{3}}[/tex]
= 16.74315781
[tex]P[\frac{(113-143)}{16.74315781} \le \frac{x-\mu\bar x}{\mu\bar x} \le \frac{(176-143}{16.74315781}][/tex]
[tex]P( -1.7918 \le X \le 1.9710 )[/tex]
[tex]P( z < 1.9710 ) - P( Z < -1.7918 )[/tex]
[tex]0.9756 - 0.0366[/tex]
= 0.9391
c) [tex]P( 113 \le \bar x\le 176 ) , n = 83[/tex]
[tex]\mu \bar x = 143[/tex]
[tex]\sigma\bar x=\frac{\sigma}{\sqrt{n} }[/tex]
= [tex]\frac{29}{\sqrt{83} }[/tex]
= 3.18316354
[tex]P[\frac(113-143){3.18316354}\le \frac {x-\mu\bar x}{\sigma \bar x} \le \frac{176-143}{3.18316354}[/tex]
[tex]P( -9.4246 \le X \le 10.3670 )[/tex]
[tex]P( z < 10.3670 ) - P( Z < -9.4246 )[/tex]
= 1.0000 - 0.0000
= 1.0000
Therefore,
a) The probability that her weight is between 113 lb and 176 lb is 0.7220.
b) The probability that they have a mean weight between 113 16 and 176 is 0.9391.
c) The probability that they have a mean weight between 113 ib and 176 lb is 1.0000.
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