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Lera25 [3.4K]
3 years ago
11

Jill's bowling scores are approximately normally distributed with mean 170 and standard deviation 25, while jack's scores are ap

proximately normally distributed with mean 160 and standard deviation 10. if jack and jill each bowl one game, then assuming their scores are independent random variables, find the probability that
Mathematics
1 answer:
olya-2409 [2.1K]3 years ago
5 0
Jill's bowling scores<span> are </span>approximately normally distributed<span> with </span>mean 170<span> and </span>standard deviation<span> 20,</span>while Jack's scores<span> are ... </span>If Jack<span> and </span>Jill each bowl one game<span>, </span>then assuming<span> that </span>their scores<span> are ... The difference of </span>means<span> of </span>Jack<span> and </span>Jill<span> follows a </span>normal distribution<span> with </span>mean<span> -</span>10<span> and </span>standard deviation<span> is: </span>25<span>.</span>
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<em><u>Solution:</u></em>

<em><u>Given system of equations are:</u></em>

y = x^2 - 6x + 12 ------ eqn\ 1\\\\y = 2x - 4 ---------- eqn\ 2

<em><u>Substitute eqn 2 in eqn 1</u></em>

x^2 - 6x + 12 = 2x - 4

Make the right side of equation 0

x^2 - 6x + 12 - 2x + 4 = 0\\\\x^2 -8x + 16 = 0

<em><u>Solve by quadratic equation</u></em>

\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=1,\:b=-8,\:c=16:\\\\x=\frac{-\left(-8\right)\pm \sqrt{\left(-8\right)^2-4\cdot \:1\cdot \:16}}{2\cdot \:1}\\\\x=\frac{-\left(-8\right)\pm \sqrt{0}}{2\cdot \:1}\\\\x = \frac{8}{2}\\\\x = 4

<em><u>Substitute x = 4 in eqn 2</u></em>

y = 2(4) - 4

y = 8 - 4

y = 4

Thus solution is (4, 4)

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