A firm report 35% of the probability that its accounts receivable from other business firms are overdue. If an accountant takes a random sample of five such accounts, then construct the probability distribution table for accounts are overdue.
Let "X=" the number of accounts which are overdue: "X\\sim Bin(n,p)"
Given "p=0.35, n=5."
"P(X=0)=\\dbinom{5}{0}0.35^0(1-0.35)^{5-0}"
"=0.1160290625"
"P(X=1)=\\dbinom{5}{1}0.35^1(1-0.35)^{5-1}"
"=0.3123859375"
"P(X=2)=\\dbinom{5}{2}0.35^2(1-0.35)^{5-2}"
"=0.336415625"
"P(X=3)=\\dbinom{5}{3}0.35^3(1-0.35)^{5-3}"
"=0.181146875"
"P(X=4)=\\dbinom{5}{4}0.35^4(1-0.35)^{5-4}"
"=0.0487703125"
"P(X=5)=\\dbinom{5}{5}0.35^5(1-0.35)^{5-5}"
"=0.0052521875"
"\\def\\arraystretch{1.5}\n \\begin{array}{c:c}\n x &p(x) \\\\ \\hline\n 0 & 0.1160290625 \\\\\n \\hdashline\n 1 & 0.3123859375 \\\\\n\\hdashline\n 2 & 0.336415625 \\\\\n\\hdashline\n 3 & 0.181146875 \\\\\n\\hdashline\n 4 & 0.0487703125 \\\\\n\\hdashline\n 5 & 0.0052521875 \\\\\n\\end{array}"
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