This table shows the proportion of nests with different numbers of eggs:
\n| Clutch Size | \n3 | \n4 | \n5 | \n6 | \n7 | \n8 | \n9 | \n
| $P(C=c)$ | \n{e3} | \n{e4} | \n{e5} | \n{e6} | \n{e7} | \n{e8} | \n{e9} | \n
a) {advicea}
\nb) We want to calculate the expected number of eggs in the clutch.
\nSince we have discrete data our formula for expected value is $E[X]=\\sum_{i=1}^nx_i\\cdot P(X=x_i)$, where:
\nHence,
\n\\begin{align}
E[X] &= \\sum_{i=1}^nx_i\\cdot P(X=x_i) \\\\
&= 3\\cdot P(X=3) + 4\\cdot P(X=4) + 5\\cdot P(X=5) + 6\\cdot P(X=6) + 7\\cdot P(X=7) + 8\\cdot P(X=8) + 9\\cdot P(X=9) \\\\
&= 3\\cdot\\var{e3} + 4\\cdot\\var{e4} + 5\\cdot\\var{e5} + 6\\cdot\\var{e6} + 7\\cdot\\var{e7} + 8\\cdot\\var{e8} + 9\\cdot\\var{e9} \\\\
&= \\var{3*e3} +\\var{4*e4} + \\var{5*e5} + \\var{6*e6} + \\var{7*e7} +\\var{8*e8} + \\var{9*e9} \\\\
&= \\var{expected}
\\end{align}
Hence, our expected number of eggs in the clutch is $\\var{round_exp}$ (rounded to two decimal places if necessary).
\nc) Our formula for variance is $Var(X) = E[X^2] - E[X]^2.$
\nWe have already calculated $E[x]$ in part b so now we must find $E[X^2]$.
\nSo,
\n\\begin{align}
E[X^2] &= \\sum_{i=1}^nx_i^2\\cdot P(X=x_i) \\\\
&= 3^2\\cdot P(X=3) + 4^2\\cdot P(X=4) + 5^2\\cdot P(X=5) + 6^2\\cdot P(X=6) + 7^2\\cdot P(X=7) + 8^2\\cdot P(X=8) + 9^2\\cdot P(X=9) \\\\
&= 9\\cdot\\var{e3} + 16\\cdot\\var{e4} + 25\\cdot\\var{e5} + 36\\cdot\\var{e6} + 49\\cdot\\var{e7} + 64\\cdot\\var{e8} + 81\\cdot\\var{e9} \\\\
&= \\var{9*e3} +\\var{16*e4} + \\var{25*e5} + \\var{36*e6} + \\var{49*e7} +\\var{64*e8} + \\var{81*e9} \\\\
&= \\var{squared}.
\\end{align}
Hence,
\n\\begin{align}
Var(X) &= E[X^2] - E[X]^2 \\\\
&= \\var{squared} - \\var{expected}^2 \\\\
&= \\var{squared} - \\var{expected^2} \\\\
&= \\var{variance}.
\\end{align}
Therefore our variance in the number of eggs is $\\var{round_var}$ (rounded to two decimal places if necessary).
\nd) We now want to find the expected number of chicks surviving.
\nOur formula for this will be $E[X]=\\sum_{i=1}^n\\sqrt{x_i}\\cdot P(X=x_i)$.
\nSo,
\n\\begin{align}
E[X] &= \\sum_{i=1}^n\\sqrt{x_i}\\cdot P(X=x_i) \\\\
&= \\sqrt{3}\\cdot P(X=3) + \\sqrt{4}\\cdot P(X=4) + \\sqrt{5}\\cdot P(X=5) + \\sqrt{6}\\cdot P(X=6) + \\sqrt{7}\\cdot P(X=7) + \\sqrt{8}\\cdot P(X=8) + \\sqrt{9}\\cdot P(X=9) \\\\
&= \\sqrt{3}\\cdot\\var{e3} + \\sqrt{4}\\cdot\\var{e4} + \\sqrt{5}\\cdot\\var{e5} + \\sqrt{6}\\cdot\\var{e6} + \\sqrt{7}\\cdot\\var{e7} + \\sqrt{8}\\cdot\\var{e8} + \\sqrt{9}\\cdot\\var{e9} \\\\
&= \\var{transform_exp}.
\\end{align}
Hence, our expected number of chicks surviving is $\\var{round_Texp}$ to two decimal places.
\n\nUse this link to find some resources which will help you revise this topic.
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", "templateType": "anything", "can_override": false}, "advice6": {"name": "advice6", "group": "Part a - probability ", "definition": "\"We want the probability that there are 6 or more eggs in the clutch, i.e., $P(C\\\\geq6)$.
\\nThis means that the clutch could contain 6, 7, 8 or 9 eggs so, to calculate $P(C\\\\geq6)$, we use the following sum:
\\n$P(C\\\\geq6)=P(C=6)+P(C=7)+P(C=8)+P(C=9)$.
\\nSo,
\\n\\\\begin{align}
P(C\\\\geq6)&=\\\\var{e6}+\\\\var{e7}+\\\\var{e8}+\\\\var{e9} \\\\\\\\
&=\\\\var{eggprob}.
\\\\end{align}
Hence, the probability that there are 6 or more eggs in the clutch is $\\\\var{eggprob}$.
\"", "description": "", "templateType": "long string", "can_override": false}, "advice7": {"name": "advice7", "group": "Part a - probability ", "definition": "\"We want the probability that there are 7 or more eggs in the clutch, i.e., $P(C\\\\geq7)$.
\\nThis means that the clutch could contain 7, 8 or 9 eggs so, to calculate $P(C\\\\geq7)$, we use the following sum:
\\n$P(C\\\\geq7)=P(C=7)+P(C=8)+P(C=9)$.
\\nSo,
\\n\\\\begin{align}
P(C\\\\geq7)&=\\\\var{e7}+\\\\var{e8}+\\\\var{e9} \\\\\\\\
&=\\\\var{eggprob}.
\\\\end{align}
Hence, the probability that there are 7 or more eggs in the clutch is $\\\\var{eggprob}$.
\"", "description": "", "templateType": "long string", "can_override": false}, "advice8": {"name": "advice8", "group": "Part a - probability ", "definition": "\"We want the probability that there are 8 or more eggs in the clutch, i.e., $P(C\\\\geq8)$.
\\nThis means that the clutch could contain 8 or 9 eggs so, to calculate $P(C\\\\geq8)$, we use the following sum:
\\n$P(C\\\\geq8)=P(C=8)+P(C=9)$.
\\nSo,
\\n\\\\begin{align}
P(C\\\\geq8)&=\\\\var{e8}+\\\\var{e9} \\\\\\\\
&=\\\\var{eggprob}.
\\\\end{align}
Hence, the probability that there are 8 or more eggs in the clutch is $\\\\var{eggprob}$.
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