A certain type of digital camera comes in either a 3-megapixel version or a 4 megapixel version. A camera store has received a shipment of 12 of these cameras, of which 5 have

3-megapixel resolution. Suppose that 4 of these cameras are randomly selected to be stored behind the counter; the other 8 are placed in the storeroom. Let X = the number of 3-megapixel cameras among the 4 selected for behind-the-counter storage.

(a) What kind of distribution does X have (name and values of all parameters) ?

(b) Compute P (X = 2), P (X? 2) and P (X? 2).

(c) Calculate the mean value and the standard deviation of X.

a) Hyper-geometric distribution

b1) P (X=2) = 0.424

b2) P (X>2) = 0.152

C1) Mean of X = 1.67

C2) Standard Deviation of X = 0.65

Step-by-step explanation:

b1) P (X = 2) = (5C2 * 7C2) / (12C4) = (10 * 21) / (495) = 210/495

P (X=2) = 0.424

b2) P (X>2) = 1 - P (X≤2) = 1 - [P (X=0) + P (X=1) + P (X=2) ]

P (X=0) = (7C4) / (12C4) = 35/495 = 0.071

P (X=1) = (5C1 * 7C3) / (12C4) = (5 * 35) / (495) = 175/495 = 0.354

P (X=2) = 0.424

P (X>2) = 1 - P (X≤2) = 1 - [0.424 + 0.071 + 0.354]

P (X>2) = 0.152

C1) Mean of X = nk/N

k = number of 3-megapixel cameras = 5

n = number of selected cameras = 4

N = Total number of cameras = 12

Mean of X = 5*4/12

Mean of X = 1.67

C2) Standard Deviation of X

Standard deviation of X = (nk/N * (1-k/N) * (N-n) / (N-1)) 1/2

Standard Deviation of X = (4 * (5/20) * (1-5/12) * ((12-4) / (12-1)) ^1/2 = 0.65

Standard Deviation of X = 0.65