Question

The University of Washington claims that it graduates 85% of its basketball players. An NCAA investigation about the graduation rate finds that of the last 20 players entering the program only 11 graduated. If the university’s claim is correct, the number of players among the 20 should follow the binomial distribution B(20, 0.85).


a) What is the probability that 11 out of the 20 would graduate? Write how the binomial formula is used to calculate the probability.

(Answer: 0.74%) ?

b) To what extent do you think the university’s claim is true?

(Answer: Very low extent) ?

c) What is the probability that all 20 athletes would have graduated? Write how the binomial formula is used to calculate the probability.

(Answer: 3.86%) ?

d) Find the mean and standard deviation of the number of graduates out of the 20 players.

Answer 1

Probabilities are used to determine the chances of events

The given parameters are:

• Sample size: n = 20
• Proportion: p = 85%

(a) What is the probability that 11 out of the 20 would graduate?

Using the binomial probability formula, we have:

$P(X = x) = ^nC_x p^x(1 - p)^{n -x}$

So, the equation becomes

$P(x = 11) = ^{20}C_{11} \times (85\%)^{11} \times (1 - 85\%)^{20 -11}$    

This gives

$P(x = 11) = 167960 \times (0.85)^{11} \times 0.15^{9}$

$P(x = 11) = 0.0011$

Express as percentage

$P(x = 11) = 0.11\%$

Hence, the probability that 11 out of the 20 would graduate is 0.11%

(b) To what extent do you think the university’s claim is true?

The probability 0.11% is less than 50%.

Hence, the extent that the university’s claim is true is very low

(c) What is the probability that all  20 would graduate?

Using the binomial probability formula, we have:

$P(X = x) = ^nC_x p^x(1 - p)^{n -x}$

So, the equation becomes

$P(x = 20) = ^{20}C_{20} \times (85\%)^{20} \times (1 - 85\%)^{20 -20}$    

This gives

$P(x = 20) = 1 \times (0.85)^{20} \times (0.15\%)^0$

$P(x = 20) = 0.0388$

Express as percentage

$P(x = 20) = 3.88\%$

Hence, the probability that all 20 would graduate is 3.88%

(d) The mean and the standard deviation

The mean is calculated as:

$\mu = np$

So, we have:

$\mu = 20 \times 85\%$

$\mu = 17$

The standard deviation is calculated as:

$\sigma = np(1 - p)$

So, we have:

$\sigma = 20 \times 85\% \times (1 - 85\%)$

$\sigma = 20 \times 0.85 \times 0.15$

$\sigma = 2.55$

Hence, the mean and the standard deviation are 17 and 2.55, respectively.

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