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Snezhnost [94]
2 years ago
10

The probability distribution for a

Mathematics
1 answer:
adell [148]2 years ago
8 0

Considering the distribution given in the table, the desired probability is:

P(-2 \leq X \leq 2) = 0.6

<h3>What is the probability distribution given in the table?</h3>

The distribution is:

  • P(X = -5) = 0.17.
  • P(X = -3) = 0.13.
  • P(X = -2) = 0.33.
  • P(X = 0) = 0.16.
  • P(X = 2) = 0.11.
  • P(X = 3) = 0.10.

The values between -2 and 2(inclusive) are -2, 0 and 2, hence the desired probability is:

P(-2 \leq X \leq 2) = P(X = -2) + P(X = 0) + P(X = 2) = 0.33 + 0.16 + 0.11 = 0.6

More can be learned about probabilities at brainly.com/question/14398287

#SPJ1

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y = \stackrel{\stackrel{m}{\downarrow }}{\cfrac{5}{6}}x+\cfrac{7}{6}\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}

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\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{\cfrac{5}{6}} ~\hfill \stackrel{reciprocal}{\cfrac{6}{5}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{6}{5}}}

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(\stackrel{x_1}{-8}~,~\stackrel{y_1}{9})\qquad\qquad \stackrel{slope}{m}\implies -\cfrac{6}{5} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{9}=\stackrel{m}{-\cfrac{6}{5}}(x-\stackrel{x_1}{(-8)})\implies y-9=-\cfrac{6}{5}(x+8) \\\\\\ y-9=-\cfrac{6}{5}x-\cfrac{48}{5}\implies y=-\cfrac{6}{5}x-\cfrac{48}{5}+9\implies y=-\cfrac{6}{5}x-\cfrac{3}{5}

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