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A researcher wishes to conduct a study of the color preferences of new car buyers. Suppose that 50% of this population prefers the color green. If 16 buyers are randomly selected, what is the probability that exactly 10 buyers would prefer green

Sagot :

MrRoyal

Answer:

[tex]P(X=10) = 0.1222[/tex]

Step-by-step explanation:

Represent Green with G

So,

[tex]G = 50\%[/tex]

Required

Determine the probability that 10 out of 16 prefer green

This question is an illustration of binomial distribution and will be solved using the following binomial distribution formula.

[tex]P(X=x) = ^nC_xG^x(1-G)^{n-x}[/tex]

In this case:

[tex]n = 16[/tex] -- number of people

[tex]x = 10[/tex] -- those that prefer green

So, the expression becomes:

[tex]P(X=10) = ^{16}C_{10}G^{10}(1-G)^{16-10}[/tex]

[tex]P(X=10) = ^{16}C_{10}G^{10}(1-G)^{6}[/tex]

Substitute 50% for G (Express as decimal)

[tex]P(X=10) = ^{16}C_{10}*0.50^{10}*(1-0.50)^{6}[/tex]

[tex]P(X=10) = ^{16}C_{10}*0.50^{10}*0.50^{6}[/tex]

Apply law of indices

[tex]P(X=10) = ^{16}C_{10}*0.50^{10+6[/tex]

[tex]P(X=10) = ^{16}C_{10}*0.50^{16[/tex]

Solve 16C10

[tex]P(X=10) = \frac{16!}{(16-10)!10!} *0.50^{16[/tex]

[tex]P(X=10) = \frac{16!}{6!10!} *0.50^{16[/tex]

[tex]P(X=10) = \frac{16*15*14*13*12*11*10!}{6!10!} *0.50^{16[/tex]

[tex]P(X=10) = \frac{16*15*14*13*12*11}{6!} *0.50^{16[/tex]

[tex]P(X=10) = \frac{16*15*14*13*12*11}{6*5*4*3*2*1} * 0.50^{16[/tex]

[tex]P(X=10) = \frac{5765760}{720} * 0.50^{16[/tex]

[tex]P(X=10) = 8008 * 0.50^{16[/tex]

[tex]P(X=10) = 8008 * 0.00001525878[/tex]

[tex]P(X=10) = 0.12219231024[/tex]

[tex]P(X=10) = 0.1222[/tex]

Hence, the required probability is 0.1222

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