Question

Solve the problem and make a mind map of all math tools you used to solve it.

A card is drawn randomly from a deck then replaced. The deck is shuffled. Ten trials are carried out.

a) Use the binomial distribution to determine the probability that there are exactly 5 diamonds in 10 trials. [ /2K, /2A, /2T, /2C]

b) Could you reasonably model this distribution using a normal approximation? Explain.

[ /2K, /2A, /2T, /2C]

c) Determine the mean and standard deviation of the normal approximation.

[ /2K, /2A, /2T, /2C]

d) Use the normal approximation to determine the probability of getting exactly 5 diamonds.

[ /2K, /2A, /2T, /2C]

e) How does the answer to part d) compare with answer to part a)? .

[ /2K, /2A, /2T, /2C]

Answer 1

We are assuming each trial as independent. The no of trials is 10, i.e. n=10.

The probability of getting a diamond in a random trial is 1/4 as there are 13 diamonds in a full deck( i,e, 52 cards), so p=0.25.

a) To get exactly 5 diamonds, we would see the no of ways that it can happen, and i.e. C510 (its by the basic combination rule of selecting say k similar items from n items)

Now probability of each of this case is,

P= P(diamond)5xP(non-diamond)5

Total probabiltiy is,

P= P(diamond)5xP(non-diamond)5x(no of ways to get it)

=(1/4)5x(3/4)5x(C510)

=0.0584

b) No, because to resonably model binomial distribution as a normal distribution, the p value should be away from 0 & 1 and the n should be large. A reasonable thumb of rule is np>5 and n(1-p)>5.

here, np = 10 x 0.25 = 2.5 < 5 and n(1-p) = 10 x 0.75 = 7.5 > 5

since one of the two conditions is not followed so its not reasonable to approximate it

c) If we approximate it to normal then, the mean and standard deviation of normal is same as the binomial

i.e. mean = µ = p = 0.25 and standard deviation = σ = √( p(1-p) ) = √( 0.25x0.75 ) = 0.433

d) The normal distribution is given as,

so to find probabilty for exactly 5 diamonds, we put x = 5,

e) the answer to d and a part are quite different, probably as we said in part b that its not reasonable to approximate these distribution, means their distributions don't resembles and so thats why the probabilty are also different here.