for the following problem, you will need the formulas below.\nthe mean/expected value of a discrete…

for the following problem, you will need the formulas below.\nthe mean/expected value of a discrete probability distribution is:\nμ = ∑x·p(x).\nthe standard deviation of a discrete probability distribution is:\nσ = √(∑x²·p(x) - μ²).\nfor the entire world’s population, approximately 17.3% of all people speak chinese as their native language. suppose four people are randomly selected from the earth’s population, and define x to be the number of these who speak chinese as their native language. the probability distribution associated with this random variable is given below with one missing probability.\nenter the missing probability below, rounded to 3 significant figures.\n\n|x|p(x)|\n|0|0.468|\n|1|0.391|\n|2|0.123|\n|3|0.017|\n|4|0.001|\n\nround the answers to the following questions to two places after the decimal.\n\n● give the mean (i.e. expected value, e(x)) of this distribution: μ = e(x) =\n● give the standard deviation of this distribution: σ =\n● give the variance (var(x)) of this distribution: var(x)=σ² =

for the following problem, you will need the formulas below.\nthe mean/expected value of a discrete probability distribution is:\nμ = ∑x·p(x).\nthe standard deviation of a discrete probability distribution is:\nσ = √(∑x²·p(x) - μ²).\nfor the entire world’s population, approximately 17.3% of all people speak chinese as their native language. suppose four people are randomly selected from the earth’s population, and define x to be the number of these who speak chinese as their native language. the probability distribution associated with this random variable is given below with one missing probability.\nenter the missing probability below, rounded to 3 significant figures.\n\n|x|p(x)|\n|0|0.468|\n|1|0.391|\n|2|0.123|\n|3|0.017|\n|4|0.001|\n\nround the answers to the following questions to two places after the decimal.\n\n● give the mean (i.e. expected value, e(x)) of this distribution: μ = e(x) =\n● give the standard deviation of this distribution: σ =\n● give the variance (var(x)) of this distribution: var(x)=σ² =

Answer

Explanation:

Step1: Recall probability property

The sum of all probabilities in a probability - distribution is 1. Let the missing probability be $p$. Then $0.468 + 0.391+0.123 + 0.017+0.001 + p=1$.

Step2: Solve for the missing probability

$p = 1-(0.468 + 0.391+0.123 + 0.017+0.001)=1 - 1=0$ (the sum of the given probabilities is already 1, so the missing probability is 0).

Step3: Calculate the mean $\mu$

$\mu=\sum_{x = 0}^{4}x\cdot P(x)=0\times0.468 + 1\times0.391+2\times0.123 + 3\times0.017+4\times0.001$ $=0 + 0.391+0.246+0.051 + 0.004=0.692\approx0.69$.

Step4: Calculate $\sum_{x = 0}^{4}x^{2}\cdot P(x)$

$\sum_{x = 0}^{4}x^{2}\cdot P(x)=0^{2}\times0.468 + 1^{2}\times0.391+2^{2}\times0.123 + 3^{2}\times0.017+4^{2}\times0.001$ $=0+0.391 + 0.492+0.153+0.016 = 1.052$.

Step5: Calculate the standard deviation $\sigma$

$\sigma=\sqrt{\sum_{x = 0}^{4}x^{2}\cdot P(x)-\mu^{2}}=\sqrt{1.052-(0.692)^{2}}=\sqrt{1.052 - 0.479664}=\sqrt{0.572336}\approx0.76$.

Step6: Calculate the variance $Var(x)$

$Var(x)=\sigma^{2}=(0.76)^{2}=0.58$.

Answer:

Missing probability: 0 Mean ($\mu$): 0.69 Standard deviation ($\sigma$): 0.76 Variance ($Var(x)$): 0.58