MATHEMATICS COLLEGE

Some parts of California are particularly earthquake- prone. Suppose that in one metropolitan area, 25% of all homeowners are insured against earthquake damage. Four homeowners are to be selected at random; let X denote the number among the four who have earthquake insurance. a. Find the probability distribution of X. [Hint: Let S denote a homeowner who has insurance and F one who does not. Then one possible outcome is SFSS, with proba bility (.25)(.75)(.25)(.25) and associated X value 3. There are 15 other outcomes.] b. Draw the corresponding probability histogram. c. What is the most likely value for X

Answers

Answer 1

Answer:

Answer:

a. Binomial random variable (n=4, p=0.25)

b. Attached.

c. X=1

Step-by-step explanation:

This can be modeled as a binomial random variable, with parameters n=4 (size of the sample) and p=0.25 (proportion of homeowners that are insured against earthquake damage).

a. The probability that X=k homeowners, from the sample of 4, have eartquake insurance is:

$P(x=k) = \dbinom{n}{k} p^(k)(1-p)^(n-k)\n\n\nP(x=k) = \dbinom{4}{k} 0.25^(0)\cdot0.75^(4)$

The sample space for X is {0,1,2,3,4}

The associated probabilties are:

$P(x=0) = \dbinom{4}{0} p^(0)(1-p)^(4)=1*1*0.3164=0.3164\n\n\nP(x=1) = \dbinom{4}{1} p^(1)(1-p)^(3)=4*0.25*0.4219=0.4219\n\n\nP(x=2) = \dbinom{4}{2} p^(2)(1-p)^(2)=6*0.0625*0.5625=0.2109\n\n\nP(x=3) = \dbinom{4}{3} p^(3)(1-p)^(1)=4*0.0156*0.75=0.0469\n\n\nP(x=4) = \dbinom{4}{4} p^(4)(1-p)^(0)=1*0.0039*1=0.0039\n\n\n$

b. The histogram is attached.

c. The most likely value for X is the expected value for X (E(X)).

Is calculated as:

$E(X)=np=4\cdot0.25=1$

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Stephanie claims that 26540is greater than 50. Do you agree or disagree? Explain.

Answers

Agree. 50 is a smaller

Plz can some one look at this

Answers

Add up the values for Craig and Sheila, which is 25 and 40 respectively

25+40 = 65

Answer: 65

The error made is that the random simulation is perfectly going to match up with the theoretical values. It never works out this way. Though as the number of trials increases, it should get closer and closer.

Which of the following shows the extraneous solution to the logarithmic equation x = -16
x = -4
x = 4
x = 16

Answers

Answer:

The correct answer option is x = 4.

Step-by-step explanation:

We are given the following logarithmic equation and we are to determine whether which of the given options shows its extraneous solution:

$log _ 7 ( 3 x ^ 3 + x ) - log _ 7 ( x ) = 2$

We can rewrite it as:

$log7[(3x^3+x)/(x) ]=2$

But we know that $log_7(49)=2$

So, $log7[(3x^3+x)/(x) ]=log_7(49)$

Cancelling the log to get:

$(3x^3+x)/(x) =49$

Further simplifying it to get:

$3x^2+1=49$

$3x^2=48$

$x^2=(48)/(3)$

$x^2=16$

x = 4

Answer:

The extraneous solution to the logarithmic equation is $x=-4$

Step-by-step explanation:

We have the equation:

$Log_(7) (3x^3+x)-Log_7(x)=2$

By properties of logarithms:

$LogA-LogB=Log((A)/(B))$

So, with the equation we have:

$Log_(7) ((3x^3+x))/(x)=2$

$Log_(7)( (3x^3+x)/(x))=2\nLog_(7)( (3x^3)/(x)+(x)/(x))=2\nLog_(7)( (3x^3)/(x)+1)=2\nLog_(7)(3x^2+1)=2$

This logarithm base is 7 and this equation is equal to 2, the number 7 passes as the base on the other side of the equation and the two as an exponent, after that we just to find x:

$7^2=(3x^2+1)\n49=3x^2+1\n49-1=3x^2\n(48)/(3) =x^2\n16=x^2$

Now, we can find x with square root

$16=x^2\n√(16) =√(x^2) \nx_1=4\nx_2=-4$

This equation has two answers because it is a quadratic equation, so with this logic the strange solution is -4

A company that manufactures toothpaste is studying five different package designs. Assuming that one design is just as likely to be selected by a consumer as any other design, what selection probability would you assign to each of the package designs? We would assign a probability of to the design 1 outcome, to design 2, to design 3, to design 4, and to design 5. In an actual experiment, 100 consumers were asked to pick the design they preferred. The following data were obtained. Design Number of Times Preferred 1 10 2 5 3 30 4 40 5 15 Do the data confirm the belief that one design is just as likely to be selected as another? Explain. Yes, the sum of the assigned probabilities is 1. No, a probability of about 0.20 would be assigned using the relative frequency method if selection is equally likely. Yes, the average of the assigned probabilities is 0.20. No, a probability of about 0.50 would be assigned using the relative frequency method if selection is equally likely.

Answers

Answer:

Correct option: "No, a probability of about 0.20 would be assigned using the relative frequency method if selection is equally likely."

Step-by-step explanation:

The assumption made is that all the 5 different packages are equally likely, i.e. the probability of selecting a package is $(1)/(5)=0.20$ .

The probability distribution is shown below.

According to the probability distribution:

The probability of a person preferring design 1 is,

$P(X=1)=0.10$

The probability of a person preferring design 2 is,

$P(X=2)=0.05$

The probability of a person preferring design 3 is,

$P(X=3)=0.30$

The probability of a person preferring design 4 is,

$P(X=4)=0.40$

The probability of a person preferring design 1 is,

$P(X=5)=0.15$

So it can be seen that the probability of preferring any of the 5 designs are not same.

Thus, the designs are not equally likely.

The correct option is "No, a probability of about 0.20 would be assigned using the relative frequency method if selection is equally likely."

The selection Probability determined using the relative frequency method do not match the assigned probabilities, suggesting that the data do not confirm the belief that one design is as likely to be selected as another.

The given data can be used to calculate the relative frequencies of each package design selected by the consumers.

To determine the selection probabilities using the relative frequency method, divide the number of times a design was preferred by the total number of consumers.

For example, for design 1, the selection probability would be 10/100 = 0.1.

Similarly, for design 2, the selection probability would be 5/100 = 0.05.

The selection probabilities for designs 3, 4, and 5 would be 0.3, 0.4, and 0.15 respectively.

Comparing these probabilities to the assigned probabilities, it can be observed that the assigned probabilities do not match the observed relative frequencies, indicating that the data do not confirm the belief that one design is just as likely to be selected as another.

Learn more about Probability here:

brainly.com/question/22962752

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A Popsicle store at the beach is comparing their monthly revenue with the sunglasses store next door. The shop owners find that their monthly revenue is positive correlated. This means that eating Popsicles causes people to buy sunglasses.True or False

Answers

Answer:

False

Step-by-step explanation:

Correlation is not causation.

It may mean that eating popsicles and buying sunglasses have a common cause: sunny days.

Median of a cumulative frequency graph

Answers

Answer:

Sorry this is a really really late reply but to find the median on the graph you need to find the mid value, so for example if the y axis goes up to 60, then the middle of the values will be 30. You go across this 30th value and find the median.

Hope this helps.

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