X is a binomial variable with $n = 4$ and $p = 0.3$ . Compute $P ( 1 \leq X \leq 3 )$ . Round your answer to four decimal places if necessary.
A) $P ( 1 \leq X \leq 3 ) = 0.3016$
B) $P ( 1 \leq X \leq 3 ) = 0.0756$
C) $P ( 1 \leq X \leq 3 ) = 0.7518$
D) $P ( 1 \leq X \leq 3 ) = 0.2646$
E) $P ( 1 \leq X \leq 3 ) = 0.4116$

Question

X is a binomial variable with $n = 4$ and $p = 0.3$ . Compute $P ( 1 \leq X \leq 3 )$ . Round your answer to four decimal places if necessary.
A)  $P ( 1 \leq X \leq 3 ) = 0.3016$
B)   $P ( 1 \leq X \leq 3 ) = 0.0756$
C)   $P ( 1 \leq X \leq 3 ) = 0.7518$
D)   $P ( 1 \leq X \leq 3 ) = 0.2646$
E)   $P ( 1 \leq X \leq 3 ) = 0.4116$

Quizplus · Accepted Answer

To find $P(1 \leq X \leq 3)$, we calculate the probability of $X$ being 1, 2, or 3 and sum these probabilities. The binomial probability formula is $P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}$, where $n$ is the number of trials, $p$ is the probability of success, and $k$ is the number of successes. For $n = 4$ and $p = 0.3$, we calculate:- $P(X = 1) = \binom{4}{1}(0.3)^1(0.7)^3 = 4 \times 0.3 \times 0.343 = 0.4116$- $P(X = 2) = \binom{4}{2}(0.3)^2(0.7)^2 = 6 \times 0.09 \times 0.49 = 0.2646$- $P(X = 3) = \binom{4}{3}(0.3)^3(0.7)^1 = 4 \times 0.027 \times 0.7 = 0.0756$Summing these probabilities gives $0.4116 + 0.2646 + 0.0756 = 0.7518$, which matches option C.

X Is a Binomial Variable With n=4n = 4n=4 And p=0.3p = 0.3p=0.3

X Is a Binomial Variable With $n = 4$ And $p = 0.3$