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

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

Question (a)

Probability of a diamond being drawn from a pack of 52 cards = 13/52 = 0.25

So p = 0.25

The given random variable follows a binomial distribution X ~ B(10, 0.25)

The number of successes x in n trails in binomial distribution = * px * (1-p)n-x

= n! / [ x! (n-x)! ] * px * (1-p)n-x

Here n = 10, p = 0.25

We need to calculate probability that there are exactly 5 diamonds in 10 trials

P(X=5) =   * 0.255 * (1-0.25)10-5

= 252 * 0.000977 * 0.237305

= 0.058399

= 0.058 rounded to 3 decimal places

So  probability that there are exactly 5 diamonds in 10 trials = 0.058

Question (b)

The random variable X~N(np,npq) follows the normal distribution where np = , Mean and npq = 2 which is square of standard deviation. This is normal approximation to binomial distribution

where q = 1 -p

Question (c)

Mean of the normal approximation = n * p

= 10 * 0.25

= 2.6

Standard deviation of the normal approximation = n * p * q

q = 1- p = 0.75

= 10 * 0.25 * 0.75

= 10 * 0.1875

= 1.875

= 1.369306

Question (d)

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

For normal approximation to binomail distribution we should apply the continuity correction rules in the abpve table

We need to find P(X=5) so it will become P(4.5 < X < 5.5) after continuity correction

P(4.5 < X < 5.5) = P(X<5.5) - P(X<4.5)

We will find the area to the left of z-score 4.5 and then subtract it from the area to the left of z-score 4.5

P(X<5.5)

Z-score = (X - ) /

= (5.5 - 2.5) / 1.369306

= 3 / 1.369306

= 2.190891

The area to the left of z-score 2.190891 will give us P(X<5.5) which is 0.9857701

P(X<.5)

Z-score = (X - ) /

= (4.5 - 2.5) / 1.369306

= 2 / 1.369306

= 1.460594

The area to the left of z-score 1.460594 will give us P(X<5.5) which is 0.9279366

P(4.5 < X < 5.5) = P(X<5.5) - P(X<4.5)

= 0.9857701 - 0.9279366

= 0.057824

= 0.058 rounded to 3 decimals

Probability of getting exactly 5 diamonds.by normal approximatio = 0.058

Question (e)

The probability of getting exactly 5 diamonds by binomial distribution in part (a) and normal approximation to binomial distribution in part(d) are approximately equal with the probability being 0.058 in both cases which is accurate upto 3 decimals