contestada

Seventeen individuals are scheduled to take a driving test at a particular DMV office on a certain day, nine of whom will be taking the test for the first time. Suppose that six of these individuals are randomly assigned to a particular examiner, and let X be the number among the six who are taking the test for the first time. (a) What kind of a distribution does X have (name and values of al parameters)? 17 hx;6, 9, 17) O h(x; 6,? 17 bx; 6, 9,17) (x; 6, 9, 17) 17 (b) Compute P(X = 4), P(X S 4), and P(X PLX = 4) 0.2851 PX S 4)-13946X RX24) -0.1096 X 4). (Round your answers to four decimal places.) (c) Calculaethe mean value and standard deviation of X. (Round your answers to three decimal places.)

Respuesta :

Answer:  

a) h(x; 6, 9, 17).

b) P[X=2] = 0.2036

P[X ≤ 2] = 0.2466

P[X ≥ 2] = 0.9570.

c) Mean  = 3.176.

Variance = 1.028.

Standard deviation = 1.014.

Step-by-step explanation:

From the given details K=6, n=9, N=-17.

We conclude that it is the hypergeometric distribution:  

a) h(x; 6, 9, 17).

b)

[tex]P[X=2]=\frac{(^{g}C_{2})^{17-9}C_{6-2}}{^{17}C_{6}\textrm{}}[/tex]

P[X=2] = 0.2036

P[X ≤ 2] = P(x=0)+ P(x=1) + P(x=2)

P[X ≤ 2] = 0.2466

P[X ≥ 2] = 1-[P(x=0)+P(x=1)]

P[X ≥ 2] = 0.9570.

c)

Mean= [tex]n\frac{K}{N}[/tex]

            = 3.176.

Variance = [tex]n\frac{K}{N}( \frac{N-K}{N})(\frac{N-n}{n-1} )[/tex]

               = 2.824 x 0.6471 x 0.5625

               = 1.028.

Standard deviation = [tex]\sqrt{1.028}[/tex] = 1.014.