Question

1. one of the drills done by the hillsborough high basketball team is for each player to see how many baskets he can make from the free - throw line in 1 minute. x = the number of baskets. round to 4 decimals.

a. if you stop by practice and watch 1 player run this drill, what is the probability that he will make 4 baskets?
b. if you stop by practice and watch 1 player run this drill, what is the probability that he will not make any basket?
c. if you watch 3 players run this drill, what is the probability that none of the 3 will make a basket?
d. what is the mean number of baskets you would expect a player to make in 1 minute?
e. what is the standard deviation and variance of number of baskets a player can make in 1 minute? standard deviation: ______ variance: ______

Explanation:

Step1: Find probability of making 4 baskets

From the table, when \(X = 4\), \(P(X)=0.0600\)

Step2: Find probability of making no baskets

When \(X = 0\), from the table \(P(X)=0.2500\)

Step3: Probability that none of 3 players make a basket

Since the events are independent, and the probability that one - player makes no basket is \(P(X = 0)=0.25\). The probability that none of 3 players make a basket is \(P=(0.25)^3=0.25\times0.25\times0.25 = 0.015625\approx0.0156\)

Step4: Calculate the mean \(\mu\)

The formula for the mean of a discrete - probability distribution is \(\mu=\sum_{x}x\cdot P(x)\).
\[

$$\begin{align*} \mu&=(0\times0.25)+(1\times0.3)+(2\times0.25)+(3\times0.1)+(4\times0.06)+(5\times0.03)+(6\times0.01)\\ &=0 + 0.3+0.5 + 0.3+0.24+0.15+0.06\\ &=1.6500 \end{align*}$$

\]

Step5: Calculate the variance \(\sigma^{2}\)

The formula for the variance of a discrete - probability distribution is \(\sigma^{2}=\sum_{x}(x - \mu)^{2}\cdot P(x)\)
\[

$$\begin{align*} (0 - 1.65)^{2}\times0.25&=( - 1.65)^{2}\times0.25=2.7225\times0.25 = 0.680625\\ (1 - 1.65)^{2}\times0.3&=( - 0.65)^{2}\times0.3 = 0.4225\times0.3=0.12675\\ (2 - 1.65)^{2}\times0.25&=(0.35)^{2}\times0.25 = 0.1225\times0.25=0.030625\\ (3 - 1.65)^{2}\times0.1&=(1.35)^{2}\times0.1 = 1.8225\times0.1=0.18225\\ (4 - 1.65)^{2}\times0.06&=(2.35)^{2}\times0.06=5.5225\times0.06 = 0.33135\\ (5 - 1.65)^{2}\times0.03&=(3.35)^{2}\times0.03 = 11.2225\times0.03=0.336675\\ (6 - 1.65)^{2}\times0.01&=(4.35)^{2}\times0.01 = 18.9225\times0.01=0.189225 \end{align*}$$

\]
\(\sigma^{2}=0.680625 + 0.12675+0.030625+0.18225+0.33135+0.336675+0.189225 = 1.8775\)

Step6: Calculate the standard deviation \(\sigma\)

The standard deviation \(\sigma=\sqrt{\sigma^{2}}\), so \(\sigma=\sqrt{1.8775}\approx1.3695\)

Answer:

a. \(0.0600\)
b. \(0.2500\)
c. \(0.0156\)
d. \(1.6500\)
e. Standard deviation: \(1.3695\), Variance: \(1.8775\)