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

If Sam is going 5mi per 63 min, then we can ADD 63+63=126 it takes 126minutes for 10mi, 126+63=189 it takes 189 minutes to go 15mi, now we DIVIDE 63/5 = 12.6 minutes to travel 1mi, so 12.6+12.6 =25.2 189+25.2=214.2 minutes to travel 17mi


214.2 minutes
3.57 hrs