Question

It is known that a certain basketball player will successfully make a free throw 91.55% of the time. Suppose that thebasketball player attempts to make 11 free throws. What is the probability that the basketball player will make at least 9 freethrows?

Answer 1

The probability has to be 0.9405.

Step-by-step explanation:

Answer 2

Probability: 0.9405

E(x) = 10.0705

σ = 0.9225

Step-by-step explanation:

Using the binomial distribution, we get that the probability is calculated as

`\begin{gathered} P(x)=nCx\cdot p^x\cdot(1-p)^(n-x) \\ \text{ Where nCx = }(n!)/(x!(n-x)!) \end{gathered}`

Where n is the number of trials, p is the probability of success, and x is the number of successful trials. In this case, p = 91.55% = 0.9155 and n = 11. Then, the probability is calculated as

`\begin{gathered} P(x)=11Cx\cdot0.9155^x\cdot(1-0.9155)^(11-x) \\ P(x)=11Cx\cdot0.9155^x\cdot0.0845^(11-x) \end{gathered}`

Now, we need to calculate the probability that the player will make 9 or more free throws. This probability is equal to

Therefore, the probability is equal to 0.9405

On the other hand, the expected value and the standard deviation are equal to

`\begin{gathered} E(x)=np \\ E(x)=(11)(0.9155) \\ E(x)=10.0705 \\ \\ \sigma=√(np(1-p)) \\ \sigma=√(11(0.9155)(1-0.9155)) \\ \sigma=0.9225 \end{gathered}`

Therefore, the answers are

Probability: 0.9405

E(x) = 10.0705

σ = 0.9225