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slavikrds [6]
3 years ago
10

According to a survey, 40% of adults watch a particular television show. Six adults are selected at random. The binomial distrib

ution below shows the probability that k out of 6 adults watch the television show.
P(k = 0) = 0.0467
P(k = 1) = 0.1866
P(k = 2) = 0.3110
P(k = 3) = 0.2765
P(k = 4) = 0.1382
P(k = 5) = 0.0369

The distribution is
a) symmetric.
b) skewed.
c) both symmetric and skewed.

The sum of the probabilities is:
Mathematics
2 answers:
iren [92.7K]3 years ago
7 0

Answer:

The distribution is  

b) skewed.

The sum of the probabilities is:

1

Step-by-step explanation:

In a binomial distribution, p represents the probability of success. Success in the sense that the event of interest happens. In the model presented, the probability of success p is 0.4 since we are informed that 40% of adults watch a particular television show.

The next quantity of significance in a binomial model is the number of independent trials, n. In our case there are 6 independent trials since we are told that 6 adults were selected at random. If we let the random variable K denote the number of adults out of the 6 who watch the television show, then K is a binomial random variable with parameters;

n = 6 and p = 0.4

A binomial distribution is only symmetric when either p is 0.5 or n is large. In the presented scenario none of this conditions is met since p is 0.4 while n is just 6 which is relatively small. Thus we conclude that the distribution is not symmetric but rather skewed.

The sum of the probabilities is any discrete probability distribution such as the bernoulli, binomial, negative binomial, poisson, or the geometric distribution is always equal to 1. That's a rule of thumb.

Sladkaya [172]3 years ago
6 0

Answer:

The distribution is Skewed.

The Sum of the Probabilities in the distribution is 1

Step-by-step explanation:

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A random variable X with a probability density function () = {^-x &gt; 0
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The solutions to the questions are

  • The probability that X is between 2 and 4 is 0.314
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<h3>Find the probability that X is between 2 and 4</h3>

The probability density function is given as:

f(x)= xe^ -x for x>0

The probability is represented as:

P(x) = \int\limits^a_b {f(x) \, dx

So, we have:

P(2 < x < 4) = \int\limits^4_2 {xe^{-x} \, dx

Using an integral calculator, we have:

P(2 < x < 4) =-(x + 1)e^{-x} |\limits^4_2

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P(2 < x < 4) =-(4 + 1)e^{-4} +(2 + 1)e^{-2}

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P(2 < x < 4) =-0.092 +0.406

Evaluate the sum

P(2 < x < 4) = 0.314

Hence, the probability that X is between 2 and 4 is 0.314

<h3>Find the probability that the value of X exceeds 3</h3>

This is represented as:

P(x > 3) = \int\limits^{\infty}_3 {xe^{-x} \, dx

Using an integral calculator, we have:

P(x > 3) =-(x + 1)e^{-x} |\limits^{\infty}_3

Expand the expression

P(x > 3) =-(\infty + 1)e^{-\infty}+(3+ 1)e^{-3}

Evaluate the expressions

P(x > 3) =0 + 0.199

Evaluate the sum

P(x > 3) = 0.199

Hence, the probability that X exceeds 3 is 0.199

<h3>Find the expected value of X</h3>

This is calculated as:

E(x) = \int\limits^a_b {x * f(x) \, dx

So, we have:

E(x) = \int\limits^{\infty}_0 {x * xe^{-x} \, dx

This gives

E(x) = \int\limits^{\infty}_0 {x^2e^{-x} \, dx

Using an integral calculator, we have:

E(x) = -(x^2+2x+2)e^{-x}|\limits^{\infty}_0

Expand the expression

E(x) = -(\infty^2+2(\infty)+2)e^{-\infty} +(0^2+2(0)+2)e^{0}

Evaluate the expressions

E(x) = 0 + 2

Evaluate

E(x) = 2

Hence, the expected value of X is 2

<h3>Find the Variance of X</h3>

This is calculated as:

V(x) = E(x^2) - (E(x))^2

Where:

E(x^2) = \int\limits^{\infty}_0 {x^2 * xe^{-x} \, dx

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E(x^2) = \int\limits^{\infty}_0 {x^3e^{-x} \, dx

Using an integral calculator, we have:

E(x^2) = -(x^3+3x^2 +6x+6)e^{-x}|\limits^{\infty}_0

Expand the expression

E(x^2) = -((\infty)^3+3(\infty)^2 +6(\infty)+6)e^{-\infty} +((0)^3+3(0)^2 +6(0)+6)e^{0}

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E(x^2) = 6

Recall that:

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So, we have:

V(x) = 6 - 2^2

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V(x) = 2

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Read more about probability density function at:

brainly.com/question/15318348

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<u>Complete question</u>

A random variable X with a probability density function f(x)= xe^ -x for x>0\\ 0& else

a. Find the probability that X is between 2 and 4

b. Find the probability that the value of X exceeds 3

c. Find the expected value of X

d. Find the Variance of X

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2 years ago
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Answer:

Step-by-step explanation:

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