5. in the current nba regular season, the captain of the new york knicks, jalen brunson, has made…

5. in the current nba regular season, the captain of the new york knicks, jalen brunson, has made approximately 40% of the three - point shots he attempted.\na) if brunson attempts 7 three - point shots, what is the probability that he makes at least 5 of those shots?\nb) how many three - point shots would you expect brunson to make out of 7 attempts?\nc) find the standard deviation of the number of three - point shots made out of a total of 7 attempts.
Answer
Explanation:
Step1: Identify the binomial - distribution parameters
The binomial - distribution formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successes, $p$ is the probability of success on a single trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$. Here, $n = 7$, $p=0.4$, and $1 - p = 0.6$.
Step2: Calculate the probability for part (a)
The probability that he makes at least 5 shots is $P(X\geq5)=P(X = 5)+P(X = 6)+P(X = 7)$. $P(X = 5)=C(7,5)\times(0.4)^{5}\times(0.6)^{2}=\frac{7!}{5!(7 - 5)!}\times(0.4)^{5}\times(0.6)^{2}=21\times0.01024\times0.36 = 0.0774144$. $P(X = 6)=C(7,6)\times(0.4)^{6}\times(0.6)^{1}=\frac{7!}{6!(7 - 6)!}\times(0.4)^{6}\times(0.6)^{1}=7\times0.004096\times0.6 = 0.0172032$. $P(X = 7)=C(7,7)\times(0.4)^{7}\times(0.6)^{0}=\frac{7!}{7!(7 - 7)!}\times(0.4)^{7}\times1=1\times0.0016384\times1 = 0.0016384$. $P(X\geq5)=0.0774144 + 0.0172032+0.0016384 = 0.096256$.
Step3: Calculate the expected value for part (b)
The expected value of a binomial - distribution is $E(X)=np$. Substituting $n = 7$ and $p = 0.4$, we get $E(X)=7\times0.4 = 2.8$.
Step4: Calculate the standard deviation for part (c)
The standard deviation of a binomial - distribution is $\sigma=\sqrt{np(1 - p)}$. Substituting $n = 7$, $p = 0.4$, and $1 - p = 0.6$, we get $\sigma=\sqrt{7\times0.4\times0.6}=\sqrt{1.68}\approx1.2961$.
Answer:
a) $0.096256$ b) $2.8$ c) $\approx1.2961$