Set up but do not solve for the appropriate particular solution yp for the differential equation y′′+4y=5xcos(2x) using the Method of Undetermined Coefficients (primes indicate derivatives with respect to x). In your answer, give undetermined coefficients as A, B, etc.

Answer:

0.62 = 62% probability that a U.S. adult selected at random thinks the occupation of firefighter has very great prestige.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

Desired outcomes:

Adults who consider the occupation of firefighter to have very great prestige. So the number of desired outcomes is [tex]D = 627[/tex]

Total outcomes.

All adults sampled. So [tex]T = 1010[/tex]

Probability:

[tex]P = \frac{D}{T} = \frac{627}{1010} = 0.62[/tex]

0.62 = 62% probability that a U.S. adult selected at random thinks the occupation of firefighter has very great prestige.

Answer:

1. H0 : p = 0.9

   H1 : p ≠ 0.9

2. The test is two tailed.

3. Reject the null hypothesis

Step-by-step explanation:

We are given that we have to test the claim that the proportion of people who own cats is significantly different than 90% at the 0.1 significance level.

So, Null Hypothesis, [tex]H_0[/tex] : p = 0.90

Alternate Hypothesis, [tex]H_1[/tex] : p [tex]\neq[/tex] 0.9

Here, the test is two tailed because we have given that to test  the claim that the proportion of people who own cats is significantly different than 90% which means it can be less than 0.90 or more than 0.90.

Now, test statistics is given by;

            [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)   , where,  n = sample size = 100

                                                       [tex]\hat p[/tex] = 0.94 (given)

So, Test statistics = [tex]\frac{0.94-0.90}{\sqrt{\frac{0.94(1-0.94)}{100} } }[/tex] = 1.68

Now, P-value = P(Z > 1.68) = 1 - P(Z <= 1.68)

                                           = 1 - 0.95352 = 0.0465 ≈ 0.05 or 5%

Now, our decision rule is that;

       If p-value < significance level  ⇒ Reject null hypothesis

       If p-value > significance level  ⇒ Accept null hypothesis

Since, here p-value is less than significance level as 0.05 < 0.1, so we have sufficient evidence to reject null hypothesis.

Therefore, we conclude that proportion of people who own cats is significantly different than 90%.

The probability that the sum of the dots on two dice will be 3 is 0.056, determined by finding the number of favorable outcomes (2) and dividing by the total number of outcomes (36).

The subject of this question is probability, specifically focusing on an experiment involving the tossing of two six-sided dice. To find the probability that the sum of the dots on the two top faces will be 3, we first need to determine the total number of possible outcomes when two dice are thrown. For one six-sided die, there are 6 possible outcomes. Therefore, when two dice are thrown, the total number of outcomes is 6 * 6 = 36.

Next, we need to find the number of outcomes where the sum of the dots would be 3. Looking at our dice, we can see that this can only happen in two ways: getting a 1 on the first die and a 2 on the second, or getting a 2 on the first die and a 1 on the second.

So, the number of favorable outcomes is 2.

Probability is defined as the number of favorable outcomes divided by the total number of outcomes. Hence, the probability of the sum of the dots being 3 when two dice are thrown is 2/36. To get this in three significant figures, we divide 2 by 36 which gives the decimal 0.0556. Therefore, the probability to three significant figures is 0.056.

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Answer:

[tex]z=0.842<\frac{a-204}{55}[/tex]

And if we solve for a we got

[tex]a=204 +0.842*55=250.31[/tex]

So the value of height that separates the bottom 80% of data from the top 20% is 250.31.  

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the room hotel rate of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(204,55)[/tex]  

Where [tex]\mu=204[/tex] and [tex]\sigma=55[/tex]

For this part we want to find a value a, such that we satisfy this condition:

[tex]P(X>a)=0.20[/tex]   (a)

[tex]P(X<a)=0.80[/tex]   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.80 of the area on the left and 0.20 of the area on the right it's z=0.842. On this case P(Z<0.842)=0.8 and P(z>0.842)=0.2

If we use condition (b) from previous we have this:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.842[/tex]  

[tex]P(z<\frac{a-\mu}{\sigma})=0.842[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]z=0.842<\frac{a-204}{55}[/tex]

And if we solve for a we got

[tex]a=204 +0.842*55=250.31[/tex]

So the value of height that separates the bottom 80% of data from the top 20% is 250.31.  

Answer:

Please find attached answer.

Step-by-step explanation:

The Naive Bayes model is likely to classify the document 'machine learning is fun' and 'christmas family fun' as 'entertainment', even when using Laplace smoothing.

The Naive Bayes model classifies texts based upon the likelihood of a particular term belonging to a given class. To find the target level for the document, you multiply the prior probability of the document being in a given class (either entertainment or education) with the likelihoods of the terms in the query document being in that class and you choose the class with the highest probability.

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Answer:

See the attached picture.

Step-by-step explanation:

See the attached picture.

To ensure compositions f ∘ g, g ∘ f, and f ∘ f make sense, certain conditions must be met regarding the number of inputs and outputs of the functions.

(a) For f ∘ g to make sense, the number of inputs of g must match the number of outputs of f. Therefore, n must equal p, and m must equal q.

(b) For g ∘ f to make sense, the number of inputs of f must match the number of outputs of g. Therefore, n must equal q, and m must equal p.

(c) For f ∘ f to make sense, the number of outputs of f must match the number of inputs of f. Therefore, n must equal m.

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Answer:

(5.216) mean = 61.71

(5.217) standard deviation = 8.88

(5.218) P(X>=60) = 0.9238

(5.129) L = 79, U = 69

(5.130) P(X> or = U) = 0.7058

Step-by-step explanation:

The table of the statistic is set up as shown in attachment.

(5.216) mean = summation of all X ÷ no of data.

mean = 432/7 = 61.71 birds

(5.217)Standard deviation = √ sum of the absolute value of difference of X from mean ÷ number of data

S = √ /X - mean/ ÷ 7

= √551.428/7

S = 8.88

(5.218) P (X> or = 60)

= P(Z> or =60 - 61.71/8.8 )

= P(Z>or= - 0.192)

= 1 - P(Z< or = 0.192)

= 1- 0.0762

= 0.9238

(5.219)the 15th percentile=15/100 × 7

15th percentile = 1.05

The value is the number in the first position and that is 79,

L= 79

85th percentile = 85/100 × 7 = 5.95

The value is the number in the 6th position, and that is 69

U = 69

5.130) P(X>or = 60)

= P(Z>or= 69 - 61.71/8.8)

= P(Z> or = 0.8208)

= 1 - P(Z< or = 0.8209)

= 1 - 0.2942

= 0.7058

Answer:

0.0241 or 2.41%

Step-by-step explanation:

Since each customer has a 17% probability of ordering decaffeinated coffee, the probability of a customer ordering a coffee with caffeine is:

[tex]P(C) = 1-0.17 = 0.83[/tex]

If customers make their decisions independently, the probability that all 20 order a coffee with caffeine is:

[tex]P(X=20)=P(C)^{20}\\P(X=20) = 0.83^{20}=0.0241=2.41\%[/tex]

The probability is 0.0241 or 2.41%.

Answer:

a. Should NOT be rejected because all the parameters are generally acceptable.

Step-by-step explanation:

Answer:

(a) The probability that the individual likes both vehicle #1  and vehicle #2 is 0.40.

(b) The value of P (A₂ | A₃) is 0.7143.

(c) The events A₂ and A₃ are not independent.

(d) The probability that an individual likes at least one of A₂ and A₃ given they did not like A₁ is 0.7333.

Step-by-step explanation:

The events are defined as follows:

A₁ = an individual like vehicle #1

A₂ = an individual like vehicle #2

A₃ = an individual like vehicle #3

The information provided is:

[tex]P(A_{1})=0.55\\P(A_{2})=0.65\\P(A_{3})=0.70\\P(A_{1}\cup A_{2})=0.80\\P(A_{2}\cap A_{3})=0.50\\P(A_{1}\cup A_{2}\cip A_{3})=0.88\\[/tex]

(a)

Compute the probability that the individual likes both vehicle #1  and vehicle #2 as follows:

[tex]P(A_{1}\cap A_{2})=P(A_{1})+P(A_{2})-P(A_{1}\cup A_{2})\\=0.55+0.65-0.80\\=0.40[/tex]

Thus, the probability that the individual likes both vehicle #1  and vehicle #2 is 0.40.

(b)

Compute the value of P (A₂ | A₃) as follows:

[tex]P(A_{2}|A_{3})=\frac{P(A_{2}\cap A_{3})}{P(A_{3}}\\=\frac{0.50}{0.70}\\=0.7143[/tex]

Thus, the value of P (A₂ | A₃) is 0.7143.

(c)

If two events X and Y are independent then,

[tex]P(X\cap Y)=P(X)\times P(Y)\\P(X|Y)=P(X)[/tex]

The value of P (A₂ ∩  A₃) is 0.50.

The product of the probabilities, P (A₂) and P (A₃) is:

[tex]P(A_{2})\times P(A_{3})=0.65\times0.70=0.455[/tex]

Thus, P (A₂ ∩ A₃) ≠ P (A₂) × P (A₃)

The value of P (A₂ | A₃) is 0.7143.

The value of P (A₂) is 0.65.

Thus, P (A₂ | A₃) ≠ P (A₂).

The events A₂ and A₃ are not independent.

(d)

Compute that probability that an individual likes at least one of A₂ and A₃ given they did not like A₁ as follows:

[tex]P(A_{2}\cup A_{3}|A_{1}^{c})=\frac{P((A_{2}\cup A_{3})\cap A_{1}^{c})}{P(A_{1}^{c})}\\=\frac{P((A_{2}\cup A_{3}\cup A_{1})-P(A_{1})}{1-P(A_{1})} \\=\frac{0.88-0.55}{1-0.55}\\=0.7333[/tex]

Thus, the probability that an individual likes at least one of A₂ and A₃ given they did not like A₁ is 0.7333.

Final answer:

The probability that the individual likes both vehicle #1 and vehicle #2 is 0.40. Events A2 and A3 are not independent as P(A2 and A3) does not equal P(A2)P(A3). If the individual did not like vehicle #1, the probability of liking at least one of the other two vehicles is 0.73.

Explanation:

Probability of Likes and Independency of Events

To find the probability that the individual likes both vehicle #1 and vehicle #2, which is P(A1 ∩ A2), we use the formula: P(A1 ∩ A2) = P(A1) + P(A2) - P(A1 ∪ A2). Given that P(A1) = 0.55, P(A2) = 0.65, and P(A1 ∪ A2) = 0.80, we can calculate P(A1 ∩ A2):

P(A1 ∩ A2) = 0.55 + 0.65 - 0.80 = 0.40.

The conditional probability P(A2|A3) represents the probability of liking vehicle #2 given that the individual already likes vehicle #3. Using the definition of conditional probability:

P(A2|A3) = P(A2 ∩ A3) / P(A3) = 0.50 / 0.70 = approximately 0.714.

To determine if A2 and A3 are independent events, we can check if P(A2 ∩ A3) equals P(A2)P(A3) which would be 0.65 * 0.70 = 0.455. Since P(A2 ∩ A3) = 0.50, which is not equal to 0.455, A2 and A3 are not independent.

Moreover, if A2 and A3 were independent, then knowing that A3 occurred would not change the probability of A2 happening. However, P(A2|A3) is not equal to P(A2), indicating their dependence. Lastly, when the individual did not like vehicle #1, the probability that they liked at least one of the other two vehicles can be found using the probability of the union of events, subtracting the probability of liking vehicle #1:

P(A2 ∪ A3 | not A1) = P(A2 ∪ A3) - P(A1) + P(A1 ∩ A2 ∩ A3) = 0.88 - 0.55 + 0.40 = 0.73.

Answer:

a. First four terms of the sequential partial sums

[tex]S_1=\frac{1}{7}, S_2=\frac{7^2-1}{6*7^2}, S_3 =\frac{7^3-1}{6*7^3}, S_4 =\frac{7^4-1}{6*7^4}[/tex]

b. The formula for Sn is option A = [tex]\frac{7^n-1}{6*7^n}[/tex]

c. Value of the series = [tex]\frac{1}{6}[/tex]

Step-by-step explanation:

a. First four terms of the sequential partial sums

[tex]\sum^{\infty}_{k=1}\\S_1=\frac{1}{7}\\ S_2=\frac{1}{7}+ \frac{1}{7^2}= \frac{8}{49} =\frac{7^2-1}{6*7^2}\\S_3=\frac{1}{7}+ \frac{1}{7^2}+\frac{1}{7^3} = \frac{57}{343} =\frac{7^3-1}{6*7^3}\\S_4=\frac{1}{7}+ \frac{1}{7^2}+\frac{1}{7^3}+\frac{1}{7^4} = \frac{400}{2401} =\frac{7^4-1}{6*7^4}[/tex]

b. Formula for Sn

The sum of n terms

[tex]S_n=\frac{1}{7}+ \frac{1}{7^2}+\frac{1}{7^3}+ \frac{1}{7^4}+.....+ \frac{1}{7^n}= \frac{7^n-1}{6*7^n}[/tex]

c. Value for the series

This can be given as the Sum of infinite terms;

[tex]S_{\infty}=\frac{1}{7}+ \frac{1}{7^2}+ \frac{1}{7^3}+ \frac{1}{7^4}+....+ \frac{1}{7^n}+ ....= \lim_{n \to \infty} \frac{7^n-1}{6*7^n}= \frac{1}{6}[/tex]

Answer:

75%

Step-by-step explanation:

Find the percentage corresponding to 42/56:  (0.75)(100%) = 75%

Answer:

75%

Step-by-step explanation:

Becauuuuuuseeee if you divide 42  by 56 it would be 0.75 and yeah. so take out the 0 and its 75%

Answer:

[tex]z=\frac{0.3 -0.45}{\sqrt{\frac{0.45(1-0.45)}{100}}}=-3.015[/tex]  

Step-by-step explanation:

Data given and notation

n=100 represent the random sample taken

[tex]\hat p=0.3[/tex] estimated proportion of interest

[tex]p_o=0.45[/tex] is the value that we want to test

[tex]\alpha[/tex] represent the significance level

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is 0.45.:  

Null hypothesis:[tex]p=0.45[/tex]  

Alternative hypothesis:[tex]p \neq 0.45[/tex]  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.3 -0.45}{\sqrt{\frac{0.45(1-0.45)}{100}}}=-3.015[/tex]  

The z-value is approximately -3.01.

To test the null hypothesis about a population proportion, you can use the z-test for proportions. The formula for the z-test statistic for proportions is:

[tex]z= \frac{p -p_0}{\frac{\sqrt{p_0(1 -p_0)}}{n} }[/tex]

​where:

p  is the sample proportion,  

p_0 is the hypothesized population proportion under the null hypothesis, n is the sample size.

In your case:

p =0.30 (sample proportion),

p_0​ =0.45 (hypothesized population proportion),

n=100 (sample size).

Now plug these values into the formula:

Calculate the values within the brackets first:

[tex]z= \frac{0.30 - 0.45}{\frac{\sqrt{0.45(1 -0.45)}}{100} }[/tex]

[tex]z= \frac{-0.15}{\frac{\sqrt{0.45(0.55)}}{100} }[/tex]

[tex]z= \frac{-0.15}{\frac{\sqrt{0.2475}}{100} }[/tex]

[tex]z= \frac{-0.15}{\sqrt{0.002475} }[/tex]

z≈−3.01

So, the z-value is approximately -3.01.

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Answer:

5.55% of broilers weigh between 1143 and 1242 grams

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 1511, \sigma = 198[/tex]

What proportion of broilers weigh between 1143 and 1242 grams?

This is the pvalue of Z when X = 1242 subtracted by the pvalue of Z when X = 1143.

X = 1242

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{1242 - 1511}{198}[/tex]

[tex]Z = -1.36[/tex]

[tex]Z = -1.36[/tex] has a pvalue of 0.0869

X = 1143

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{1143 - 1511}{198}[/tex]

[tex]Z = -1.36[/tex]

[tex]Z = -1.86[/tex] has a pvalue of 0.0314

0.0869 - 0.0314 = 0.0555

5.55% of broilers weigh between 1143 and 1242 grams

Answer:

No, there is not enough evidence at the 0.02 level to support the executive's claim.

Step-by-step explanation:

We are given that a publisher reports that 34% of their readers own a particular make of car. A random sample of 220 found that 30% of the readers owned a particular make of car.

And, a marketing executive wants to test the claim that the percentage is actually different from the reported percentage, i.e;

Null Hypothesis, [tex]H_0[/tex] : p = 0.34 {means that the percentage of readers who own a particular make of car is same as reported 34%}

Alternate Hypothesis, [tex]H_1[/tex] : p [tex]\neq[/tex] 0.34 {means that the percentage of readers who own a particular make of car is different from the reported 34%}

The test statistics we will use here is;

                T.S. = [tex]\frac{\hat p -p}{\sqrt{\frac{\hat p(1- \hat p)}{n} } }[/tex] ~ N(0,1)

where, p = actual % of readers who own a particular make of car = 0.34

            [tex]\hat p[/tex] = percentage of readers who own a particular make of car in a

                  sample of 220 = 0.30

            n = sample size = 220

So, Test statistics = [tex]\frac{0.30 -0.34}{\sqrt{\frac{0.30(1- 0.30)}{220} } }[/tex]

                             = -1.30

Now, at 0.02 significance level, the z table gives critical value of -2.3263 to 2.3263. Since our test statistics lie in the range of critical values which means it doesn't lie in the rejection region, so we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that the actual percentage of readers who own a particular make of car is same as reported percentage and the executive's claim that it is different is not supported.

Answer:

An insurance company reports that 75% of its claims are settled within two months of being filed. In order to test that the percent is less than seventy-five, a state insurance commission randomly selects 35 claims and determines that 23 of the 35 were settled within two months.

a) Write out the hypotheses.

Fewer than 75% of claims settled within two months of filing.

b) Calculate the test statistic.

Test statistic = percent of claims settled in two months = 23/35 = 65.7%

c) Find the p-value.

we need to use the z-score with a table

=  

standard deviation = s = \sqrt{\frac{1}{N-1} \sum_{i=1}^N (x_i - \overline{x})^2}, =

√(1/34) [(0.657)(12) + (0.343)(23)] = √0.463911765 = 0.681

let 1 = "solved" and 0 = "unresolved"

thus our mean is 0.657

z = (0.75 - 0.657)/(0.681) - 0.137

p = 0.55172

d) Do we reject the null hypothesis? Explain.

Yes, because our p value is not below 0.05 and is not substantial to prove our null hypothesis

e) What can you conclude based on this evidence?

That further testing is needed.

Let the width = X

Then the length would be 2X ( twice the width).

Area is Length x width, so you have X * 2X = 18

x *2x = 2x^2

Now you have 2x^2 = 18

Divide both sides by 2:

x^2 = 9

Take the square root of both sides:

x = sqrt(9)

x = 3

The width is 3 inches.

The width of the rectangle is determined by establishing an equation from the given area (18 square inches) and the fact that the length is twice the width, which leads to the solution that the width is 3 inches.

To find the width of a rectangle when the area is 18 square inches, and the length is twice the width, we need to set up an equation. Let the width be w inches. Then the length would be 2w inches, since it's twice the width. We know that the area (A) of a rectangle is the multiplication of its length and width, so we have:

A = length  imes width

18 = 2w  imes w

18 = 2w2

Now we solve for w:

w2 = 18 / 2

w2 = 9

w = 3

The width of the rectangle is 3 inches.

Answer:

The percent of the people who tested positive actually have the disease is 38.64%.

Step-by-step explanation:

Denote the events as follows:

X = a person has the disease

P = the test result is positive

N = the test result is negative

Given:

[tex]P(X)=0.01\\P(P|X^{c})=0.15\\P(N|X)=0.10[/tex]

Compute the value of P (P|X) as follows:

[tex]P(P|X)=1-P(P|X^{c})=1-0.15=0.85[/tex]

Compute the probability of a positive test result as follows:

[tex]P(P)=P(P|X)P(X)+P(P|X^{c})P(X^{c})\\=(0.85\times0.10)+(0.15\times0.90)\\=0.22[/tex]

Compute the probability of a person having the disease given that he/she was tested positive as follows:

[tex]P(X|P)=\frac{P(P|X)P(X)}{P(P)}=\frac{0.85\times0.10}{0.22} =0.3864[/tex]

The percentage of people having the disease given that he/she was tested positive is, 0.3864 × 100 = 38.64%.

Answer:

At 0.05 level of significance, there is no difference in the mean number of larvae per stem. This supports the scientist's claim.

(d) H0: MeanMalathion - MeanNoMalathion equals 0

H1: MeanMalathion - MeanNoMalathion not equals 0

Step-by-step explanation:

Test statistic (t) = (mean 1 - mean 2) ÷ sqrt[pooled variance (1/n1 + 1/n2)]

Let the difference between the two means be x and the pooled variance be y

n1 = 5, n2 = 12

t = x ÷ sqrt[y(1/5 + 1/12)] = x ÷ sqrt(0.283y) = x ÷ 0.532√y = 1.88x/√y

Assuming the ratio of x to √y is 0.5

t = 1.88×0.5 = 0.94

n1 + n2 = 5 + 12 = 17

degree of freedom = n1 + n2 - 2 = 17 - 2 = 15

significance level = 0.05 = 5%

critical value corresponding to 15 degrees of freedom and 5% confidence interval is 2.131

The test is a two-tailed test because the alternate hypothesis is expressed using not equal to.

The region of no rejection of the null hypothesis lies between -2.131 and 2.131

Conclusion

Fail to reject the null hypothesis because the test statistic 0.94 falls within the region bounded by the critical values.

The scientist's claim is right.

A null hypothesis is a statement from a population parameter which is either rejected or accepted (fail to reject) upon testing. It is expressed using the equality sign.

An alternate hypothesis is also a statement from a population parameter which negates the null hypothesis and is accepted if the null hypothesis is rejected. It is expressed using any of the inequality signs.

Answer:

Step-by-step explanation:

Given that an urn contains five balls numbered 1 to 5

Two balls are drawn simultaneously.

a) X = the larger of two numbers drawn

Assuming balls are drawn without replacement, (since simultaneously drawn)

the sample space would be (1,2) (1,3)(1,4)(1,5) (2,3)(2,4)(2,5) (3,4) (3,5) (4,5)

n(S) = 10

n(x=2) = 1,

n(x=3) =2

n(x=4) = 3

n(x=5) = 4

Larger value  X can take values only as 2,3,4 or 5

X          2              3            4               5

p          0.1           0.2       0.3             0.4

-------------------------------------------

b) V = sum of numbers drawn

V can take values as 3, 4, 5, 6, 7, 8 or 9

V                  3        4             5          6           7             8          9

p                  0.1     0.1         0.2      0.2         0.2        0.1        0.1

The probability that the larger number drawn will be k follows a specific probability distribution, as does the sum of the two numbers drawn. Using the given information and principles of probability, we can determine the probabilities for each value of k for both X and V. For X, the probability distribution pX(k) is [1/5, 1/5, 3/10, 1/5, 1/5]. For V, the probability distribution pV(k) is [1/20, 1/10, 3/20, 1/10, 1/10, 1/10, 3/20, 1/10, 1/20].

(a) Let's consider each possible outcome to find pX(k), the probability that the larger number drawn will be k:

If k is 1, there is only one possible outcome: drawing (1, anything). The probability of this outcome is (1/5) * (4/4) = 1/5.

If k is 2, there are two possible outcomes: drawing (2, 1) or (2, 3). The probability of these outcomes is (1/5) * (1/4) + (1/5) * (1/4) = 1/10 + 1/10 = 1/5.

If k is 3, there are three possible outcomes: drawing (3, 1), (3, 2), or (3, 4). The probability of these outcomes is (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (2/4) = 1/10 + 1/10 + 1/10 = 3/10.

If k is 4, there are four possible outcomes: drawing (4, 1), (4, 2), (4, 3), or (4, 5). The probability of these outcomes is (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (3/4) = 1/10 + 1/10 + 1/10 + 3/20 = 1/5.

If k is 5, there are five possible outcomes: drawing (5, 1), (5, 2), (5, 3), (5, 4), or (5, 5). The probability of these outcomes is (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (4/4) = 1/10 + 1/10 + 1/10 + 1/10 + 1/5 = 1/5.

Therefore, the probability distribution pX(k) is:

pX(1) = 1/5

pX(2) = 1/5

pX(3) = 3/10

pX(4) = 1/5

pX(5) = 1/5

(b) Let's consider each possible outcome to find pV(k), the probability that the sum of the two numbers drawn will be k:

If k is 2, there is only one possible outcome: drawing (1, 1). The probability of this outcome is (1/5) * (1/4) = 1/20.

If k is 3, there are two possible outcomes: drawing (1, 2) or (2, 1). The probability of these outcomes is (1/5) * (1/4) + (1/5) * (1/4) = 1/20 + 1/20 = 1/10.

If k is 4, there are three possible outcomes: drawing (1, 3), (2, 2), or (3, 1). The probability of these outcomes is (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) = 1/20 + 1/20 + 1/20 = 3/20.

If k is 5, there are four possible outcomes: drawing (1, 4), (2, 3), (3, 2), or (4, 1). The probability of these outcomes is (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) = 1/20 + 1/20 + 1/20 + 1/20 = 1/10.

If k is 6, there are five possible outcomes: drawing (1, 5), (2, 4), (3, 3), (4, 2), or (5, 1). The probability of these outcomes is (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) = 1/20 + 1/20 + 1/20 + 1/20 + 1/20 = 1/10.

If k is 7, there are four possible outcomes: drawing (2, 5), (3, 4), (4, 3), or (5, 2). The probability of these outcomes is (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) = 1/20 + 1/20 + 1/20 + 1/20 = 1/10.

If k is 8, there are three possible outcomes: drawing (3, 5), (4, 4), or (5, 3). The probability of these outcomes is (1/5) * (1/4) + (1/5) * (1/4) + (1/5) * (1/4) = 1/20 + 1/20 + 1/20 = 3/20.

If k is 9, there are two possible outcomes: drawing (4, 5) or (5, 4). The probability of these outcomes is (1/5) * (1/4) + (1/5) * (1/4) = 1/20 + 1/20 = 1/10.

If k is 10, there is only one possible outcome: drawing (5, 5). The probability of this outcome is (1/5) * (1/4) = 1/20.

Therefore, the probability distribution pV(k) is:

pV(2) = 1/20

pV(3) = 1/10

pV(4) = 3/20

pV(5) = 1/10

pV(6) = 1/10

pV(7) = 1/10

pV(8) = 3/20

pV(9) = 1/10

pV(10) = 1/20

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Answer:

Step-by-step explanation:

Hello!

X: nitrate concentration in the city's drinking water supply.

X~N(μ;δ²)

A concentration above 10 ppm is a health risk for infants less than 6 months old.

The reported nitrate concentration is between 1.59 and 2.52 ppm, values outside this range are considered unusual.

If any value of a normal distribution that is μ ± 2δ is considered unusual, it is determined that:

μ - 2δ= 1.59

μ + 2δ= 2.52

20) and 21)

I'll clear the values of the mean and the standard deviation using the given information:

a) μ - 2δ= 1.59 ⇒ μ = 1.59 + 2δ

b) μ + 2δ= 2.52 ⇒ Replace the value of Mu from "a" in "b" and clear the standard deviation:

(1.59 + 2δ) + 2δ= 2.52

1.59 + 4δ= 2.52

4δ= 2.52 - 1.59

δ= 0.93/4

δ= 0.2325

Then the value of the mean is μ = 1.59 + 2(0.2325)

μ = 1.59 + 0.465

μ = 2.055

22)

Acording to the empirical rule, 68% of a normal distribution is between μ±δ

Then you can expect 68% of the distribution between:

μ-δ= 2.055-0.2325= 1.8225

μ+δ= 2.055+0.2325= 2.2875

Correct option: C. 1.823 and 2.288 ppm

I hope it helps!

Answer:

(a) 4 possible outcomes

(b) Not possible, testing stops upon finding the bad battery,

Step-by-step explanation:

Let G denote a good battery and B denote a bad battery.

Whenever a bad battery is found, the batteries stop being tested.

(a) The outcomes for this experiment are:

B

GB

GGB

GGGB

There are 4 possible outcomes.

(b) The outcome GBGG is not possible since testing stops upon finding the bad battery, the outcome in this case would be GB.

Answer:

750,0000

Step-by-step explanation:

Answer:16.5

Step-by-step explanation:

Answer:

11 7/10

Step-by-step explanation:

Lets make the fractions into common denominators, by mulitplying 4 by 5 and 5 by 4 (see below) and follow PEMDAS

3 1/4(5) +(3 1/4(5) + 5 1/5(4))

3 5/20+ (3 5/20+5 4/20)

Simplify like terms

3 5/20+ (8 9/20)

11 14/20

Simplify fraction

11 7/10

Answer:

P(X = 17) = 0.3002

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

In this problem we have that:

[tex]n = 18, p = 0.9[/tex]

We want P(X = 17). So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 17) = C_{18,17}.(0.9)^{17}.(0.1)^{1} = 0.3002[/tex]

P(X = 17) = 0.3002

Answer:

P(X=17) = 0.3002 .

Step-by-step explanation:

We are given that the random variable X has a binomial distribution with the given probability of obtaining a success.

The above situation can be represented through Binomial distribution;

[tex]P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....[/tex]

where, n = number of trials (samples) taken = 18

            r = number of success = 17

           p = probability of success which in our question is given as 0.9 .

So, X ~ [tex]Binom(n=18,p=0.9)[/tex]

We have to find the probability of P(X = 17);

P(X = 17) = [tex]\binom{18}{17}0.9^{17} (1-0.9)^{18-17}[/tex]

               = [tex]18 \times 0.9^{17} \times 0.1^{1}[/tex]    { [tex]\because \binom{n}{r} = \frac{n!}{r! \times (n-r)!}[/tex] }

               = 0.3002

Therefore, P(X=17) = 0.3002 .

Answer:

[tex]t=1.60\ sec[/tex]

Step-by-step explanation:

Functions

We are given the function that relates the height of the ball and the time elapsed t. The function is:

[tex]h = 5 + 25t - 16t^2[/tex]

We need to know the time needed for the height to reach 4 feet, that is h=4

[tex]5 + 25t - 16t^2=4[/tex]

Rearranging

[tex]16t^2-25t -1=0[/tex]

Solving for t

[tex]\boxed{t=1.60\ sec}[/tex]

The other solution is negative and is discarded because t cannot be negative.

Answer:

The baby must weigh approximately 8.31 pounds.      

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 7.5 pounds

Standard Deviation, σ = 1.2 pounds

We are given that the distribution of weights of newborn babies is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

We have to find the third quartile or [tex]Q_3[/tex]

We have to find the value of x such that the probability is 0.75

[tex]P( X < x) = P( z < \displaystyle\frac{x - 7.5}{1.2})=0.75[/tex]  

Calculation the value from standard normal z table, we have,  

[tex]\displaystyle\frac{x - 7.5}{1.2} = 0.674\\\\x = 8.3088 \approx 8.31[/tex]

Thus, the baby must weigh approximately 8.31 pounds.

Answer:

Option C.  failing to reject a false null hypothesis

Step-by-step explanation:

Type II error states that Probability of accepting null hypothesis given the fact that null hypothesis is false.

This is considered to be the most important error.

So, from the option given to us option C matches that failing to reject a false null hypothesis.

Answer:

0.2605, 0.2188, 1.33, 4, 1.0540, 1.4142

Step-by-step explanation:

A fair die is rolled 8 times.  

a. What is the probability that the die comes up 6 exactly twice?  

b. What is the probability that the die comes up an odd number exactly five times?  

c. Find the mean number of times a 6 comes up.  

d. Find the mean number of times an odd number comes up.  

e. Find the standard deviation of the number of times a 6 comes up.  

f. Find the standard deviation of the number of times an odd number comes up.

a. A die is rolled 8 times. If A represent the number of times a 6 comes up. For a fair die the probability that the die comes up 6 is 1/6 - Thus A ~ Bin(8, 1/6)

The probability mass function  of the random variable A is  

[tex]p(A) = \left \{ {\frac{8!}{x!(8 - x)!}*(\frac{1}{6} )^{A}*(\frac{5}{6} )^{8-A} } \right. for A=0,1, ...8[/tex]

hence, p(6 twice) implies P(A=2)

that is P(2) substitute A = 2

[tex]p(2) = \left \{ {\frac{8!}{2!(8 - 2)!}*(\frac{1}{6} )^{2}*(\frac{5}{6} )^{8-2} } \right. for A=0,1, ...8[/tex]

[tex]p(2)=\frac{8!}{2!6!} *(\frac{1}{6} )^{2} *(\frac{5}{6} )^{6}[/tex]  

p(2) = 0.2605  

b. If B represent the number of times an odd number comes up. For the fair die the probability that an odd number comes up is 0.5.

Thus B ~ Bin(8, 1/2 )

The probability mass function of the random variable B is given by

[tex]p(B) = \left \{ {\frac{8!}{B!(8 - B)!}*(\frac{1}{2} )^{B\\}*(\frac{1}{2} )^{8-B} } \right. for B=0,1, ...8[/tex]

hence p(odd comes up 5 times) is

[tex]p(x=5) = p(2)=\frac{8!}{5!3!} *(\frac{1}{2} )^{5} *(\frac{1}{2} )^{3}[/tex]

p(5) = 0.2188

c. let the mean no of times a 6 comes up be μₐ

   and let the total number of outcomes be n

   using the formula μₐ = nρₐ

   μₐ = 8 * 1/6

        = 1.33

d. let the mean nos of times an odd nos comes up be μₓ

   let the total outcomes be n = 8

   let the probability odd be pb = 1/2

   μₓ = npb

        = 8 * (1/2)

        = 4

e. the standard deviation of a random variable A is given as follows

σₐ [tex]= \sqrt{np(1-p)}[/tex]

where p = 1/6 (prob 6 outcome)

n = total outcomes = 8

  [tex]= \sqrt{8*\frac{1}{6}*\frac{5}{6} }[/tex]

  = 1.0540

f. the standard dev of the binomial random variable Y is given by

σ [tex]= \sqrt{np(1-p)}[/tex]

where p = 1/2 and n = 8

  =  [tex]\sqrt{8*\frac{1}{2} *\frac{1}{2} }[/tex]

  = 1.4142

Given that the equation of the line is [tex]y=4x-3[/tex]

We need to determine the point (2,7) lies on the line.

For a point to lie on the line, substituting the point in the equation of the line makes it valid.

Thus, substituting the point (2,7) in the equation of the line [tex]y=4x-3[/tex], we get;

[tex]7=4(2)-3[/tex]

[tex]7=8-3[/tex]

[tex]7\neq 5[/tex]

Thus, the both sides of the equation are not equal.

Substituting the point (2,7) in the equation makes the equation invalid.

Thus, the point (2,7) does not lie on the line  [tex]y=4x-3[/tex]