Question

For the following question, use an appropriate normal distribution to approximate the resulting binomial distribution. Use the table in Appendix D or your calculator to find numerical values. A basketball player has a 80% chance of making a free throw. (Assume that the throws are independent of each other.) What is the probability of her making 6 or fewer free throws in 20 trials?

Answer 1

Solution:

Given that,

P = 0.80

1 - P = 0.20

n = 20

Here, $X\sim$ BIN ( n , P ) that is , BIN (20 , 0.80)

According to normal approximation binomial,

X $\rightarrow$ Normal

Mean = $\mu$ = n*P = 16

Standard deviation = $\sigma$ =$\sqrt{}$n*p*(1-p) = $\sqrt{}$ 3.2

We using continuity correction factor

P( X $\leq$ a ) = P(X < a + 0.5)

P(x < 6.5) = P((x - $\mu$ ) / $\sigma$ < (6.5 - 16) / $\sqrt{}$ 3.2)

= P(z < -5.31)

Probability = 0.0000