P(60> endobj xref 114 26 0000000016 00000 n 0000000889 00000 n 0000000946 00000 n 0000001752 00000 n 0000001910 00000 n 0000002118 00000 n 0000002444 00000 n 0000002553 00000 n 0000003631 00000 n 0000003738 00000 n 0000004124 00000 n 0000005202 00000 n 0000005314 00000 n 0000006652 00000 n 0000008443 00000 n 0000008552 00000 n 0000008929 00000 n 0000009993 00000 n 0000011487 00000 n 0000011597 00000 n 0000011675 00000 n 0000011754 00000 n 0000017445 00000 n 0000020678 00000 n 0000001091 00000 n 0000001730 00000 n trailer << /Size 140 /Info 113 0 R /Encrypt 116 0 R /Root 115 0 R /Prev 269997 /ID[<0326bc51909cfb6e47a619be98a11282><0326bc51909cfb6e47a619be98a11282>] >> startxref 0 %%EOF 115 0 obj << /Type /Catalog /Pages 111 0 R >> endobj 116 0 obj << /Filter /Standard /V 1 /R 2 /O (�@Z��ۅ� ��~$$�=�>��F��) /U (鼓�dD���A�{|�\n�4c�\\�%i|�p) /P 65476 >> endobj 138 0 obj << /S 664 /Filter /FlateDecode /Length 139 0 R >> stream It is also known as rectangular distribution. Let’s suppose a coin was tossed twice and we have to show the probability distribution of showing heads. In any normal or bell-shaped distribution, roughly... Use the normal table to validate the empirical rule. Then, go across that row until under the "0.07" in the top row. That is \alpha=2500 and \beta=4500, The probability density function of X is a. To find the probability, we need to first find the Z-scores: \(z=\dfrac{x-\mu}{\sigma}$$, For $$x=60$$, we get $$z=\dfrac{60-70}{13}=-0.77$$, For $$x=90$$, we get $$z=\dfrac{90-70}{13}=1.54$$, \begin{align*} & = \frac{1}{11} \big[x \big]_1^8\\ The intersection of the columns and rows in the table gives the probability.

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