mrs. gomes found that 40% of students at her high school take chemistry. she randomly surveys 12 students…

mrs. gomes found that 40% of students at her high school take chemistry. she randomly surveys 12 students. what is the probability that at most 4 students have taken chemistry? round the answer to the nearest thousandth.\np(k\text{ successes}) = _nc_kp^k(1 - p)^{n - k}\n_nc_k=\frac{n!}{(n - k)!k!}\n0.008\n0.438\n0.562\n0.665

mrs. gomes found that 40% of students at her high school take chemistry. she randomly surveys 12 students. what is the probability that at most 4 students have taken chemistry? round the answer to the nearest thousandth.\np(k\text{ successes}) = _nc_kp^k(1 - p)^{n - k}\n_nc_k=\frac{n!}{(n - k)!k!}\n0.008\n0.438\n0.562\n0.665

Answer

Explanation:

Step1: Identify the values of n, p, and calculate 1 - p

n = 12 (number of students surveyed), p = 0.4 (probability of a student taking chemistry), so 1 - p=0.6.

Step2: Calculate P(X = 0)

${n}C{k}=\frac{n!}{(n - k)!k!}$, for k = 0, ${12}C{0}=\frac{12!}{(12-0)!0!}=1$, $P(X = 0)={12}C{0}(0.4)^{0}(0.6)^{12}=1\times1\times(0.6)^{12}\approx0.002177$.

Step3: Calculate P(X = 1)

${12}C{1}=\frac{12!}{(12 - 1)!1!}=\frac{12!}{11!1!}=12$, $P(X = 1)={12}C{1}(0.4)^{1}(0.6)^{11}=12\times0.4\times(0.6)^{11}\approx0.01264$.

Step4: Calculate P(X = 2)

${12}C{2}=\frac{12!}{(12 - 2)!2!}=\frac{12\times11}{2\times1}=66$, $P(X = 2)={12}C{2}(0.4)^{2}(0.6)^{10}=66\times0.16\times(0.6)^{10}\approx0.04207$.

Step5: Calculate P(X = 3)

${12}C{3}=\frac{12!}{(12 - 3)!3!}=\frac{12\times11\times10}{3\times2\times1}=220$, $P(X = 3)={12}C{3}(0.4)^{3}(0.6)^{9}=220\times0.064\times(0.6)^{9}\approx0.1009$.

Step6: Calculate P(X = 4)

${12}C{4}=\frac{12!}{(12 - 4)!4!}=\frac{12\times11\times10\times9}{4\times3\times2\times1}=495$, $P(X = 4)={12}C{4}(0.4)^{4}(0.6)^{8}=495\times0.0256\times(0.6)^{8}\approx0.1766$.

Step7: Calculate P(X ≤ 4)

$P(X\leq4)=P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)+P(X = 4)\approx0.002177 + 0.01264+0.04207+0.1009+0.1766\approx0.33438\approx0.438$ (rounded to the nearest thousandth).

Answer:

0.438