Question

approximately 11% of all people are left - handed. consider 17 randomly selected people.
a) state the random variable.
rv x = the number of 17 randomly selected people that are left - handed
b) list the given numbers with correct symbols.
? = 17
p = 0.11
c) explain why this is a binomial experiment. check all that apply.
p = 11% remains constant from one randomly selected person to another
there is not a fixed number of people
there are only two outcomes for each person
there are more than two outcomes for each person
there are a fixed number of people, 17
whether or not one randomly selected person is left - handed will affect whether or not another randomly selected person is left - handed
whether or not one randomly selected person is left - handed will not affect whether or not another randomly selected person is left - handed
find the probability, to 4 decimal places:
it is possible when rounded that a probability is 0.0000
d) exactly none are left - handed.
0.1216
e) exactly 10 are left - handed.

Explanation:

Step1: Recall binomial probability formula

The binomial probability 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 $C(n,k)=\frac{n!}{k!(n - k)!}$.

Step2: Identify values for part d

For part d, we want to find the probability that exactly none are left - handed. Here, $n = 17$, $k = 0$, and $p=0.11$. First, calculate the combination $C(17,0)=\frac{17!}{0!(17 - 0)!}=1$. Then, $(1 - p)=1 - 0.11 = 0.89$.

Step3: Calculate probability for part d

Using the binomial formula $P(X = 0)=C(17,0)\times(0.11)^{0}\times(0.89)^{17}$. Since any non - zero number to the power of 0 is 1, $(0.11)^{0}=1$. So $P(X = 0)=1\times1\times(0.89)^{17}\approx0.1402$.

Step4: Identify values for part e

For part e, we want to find the probability that exactly 10 are left - handed. Here, $n = 17$, $k = 10$, and $p = 0.11$. Calculate the combination $C(17,10)=\frac{17!}{10!(17 - 10)!}=\frac{17!}{10!7!}=\frac{17\times16\times15\times14\times13\times12\times11}{7\times6\times5\times4\times3\times2\times1}=19448$. And $(1 - p)=0.89$.

Step5: Calculate probability for part e

Using the binomial formula $P(X = 10)=C(17,10)\times(0.11)^{10}\times(0.89)^{7}$.
$(0.11)^{10}\approx2.5937\times10^{- 10}$, $(0.89)^{7}\approx0.4422$. Then $P(X = 10)=19448\times2.5937\times10^{-10}\times0.4422\approx2.25\times10^{-6}$.

Answer:

d) $0.1402$
e) $0.0000$ (rounded to 4 decimal places)