Question

If 60% of all women are employed outside the home find the probability that in a sample of 20 women at least 10 are employed

Answer 1

SOLUTION:

From given data,

If 60% of all women are employed outside the home find the probability that in a sample of 20 women at least 10 are employed

P(X > 10), n= 20. P= 0.6

P(X > 10)P(10< X 20)P(X 10, 11,12, 13, 14, 15, 16,17,18,19,20)

19) + P(X = 20)(because these are disjoint events)

  

PlX = 10)-( 10) (0.6)10(1-0.6)10-184756(0.6)10(0.4)10 )(0.6)10(1-0.6)"-184756(0.6)"(0.4)10 )

ะ 0.117142

$P(X =11) =\binom{20}{11}(0.6)^{11}(1-0.6)^{9} = 167960(0.6)^{11}(0.4)^{9}$

0.159738

$P(X =12) =\binom{20}{12}(0.6)^{12}(1-0.6)^{8} = 125970(0.6)^{12}(0.4)^{8}$

0.179706

20 P(X = 13)-(B (0.6)13 (1-0.6). 77520(0.6)13(0.4)7

  $\approx 0.165882$

$P(X =14) =\binom{20}{14}(0.6)^{14}(1-0.6)^{6} = 38760(0.6)^{14}(0.4)^{6}$

  $\approx 0.124412$

$P(X =15) =\binom{20}{15}(0.6)^{15}(1-0.6)^{5} = 15504(0.6)^{15}(0.4)^{5}$

  $\approx 0.074647$

$P(X =16) =\binom{20}{16}(0.6)^{16}(1-0.6)^{4} = 4845(0.6)^{16}(0.4)^{4}$

$\approx 0.034991$

$P(X =17) =\binom{20}{17}(0.6)^{17}(1-0.6)^{3} = 1141(0.6)^{17}(0.4)^{3}$

$\approx 0.01235$

$P(X =18) =\binom{20}{18}(0.6)^{18}(1-0.6)^{2} = 190(0.6)^{18}(0.4)^{2}$

  $\approx 0.003087$

$P(X =19) =\binom{20}{19}(0.6)^{19}(1-0.6)^{1} = 20(0.6)^{19}(0.4)^{1}$

  $\approx 0.004875$

$P(X =20) =\binom{20}{20}(0.6)^{20}(1-0.6)^{0} = 1(0.6)^{20}(0.4)^{0}$

0,00003656

$=0.872479$

$\approx 0.872$

P(X > 10) 0.872

Using excel function 1 - BinomDist(9.20,0.6 true) or TI-83/84 function 1-binomdf(20,0.6,9),exact answer is 0.8724787539

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