An insurance company offers two types of insurance: health insurance and life insurance. Let X be the annual claim on the health insurance and Y be the annual claim on the life insurance. The joint density function of X and Y is: Find the probability that the total claims exceed 0.5 but is less than 2 and the claim on the life insurance is less than 1.5.
Answers
The joint density equation is missing in the question. The equation is [tex]f(x,y)=\left\{\begin{matrix}2x^3, 0\leq x\leq 1,0\leq y\leq 2\\ 0, \text{ otherwise}\end{matrix}\right.[/tex]
Probability is 0.59
Step-by-step explanation:
We have
x -- denotes the annual claim health insurance
y -- denotes the annual claim life insurance
And joint density of x and y is given to us
[tex]f(x,y)=\left\{\begin{matrix}2x^3, 0\leq x\leq 1,0\leq y\leq 2\\ 0, \text{ otherwise}\end{matrix}\right.[/tex]
Event that the claim exceeds 0.5 but less than 2 can be written as 0.5≤ x+y ≤2
and claim of life insurance less than 1.5 can be written as 0 ≤ y ≤ 1.5
Required probability
= P( 0.5≤ x+y ≤2 ∩ 0 ≤ y ≤ 1.5 )
Lets plot this, to find the common region
For common region
when 0≤ x ≤ 0.5, then 0.5 ≤ y ≤ 1.5 ---- for region I
when 0.5≤ x ≤ 1, then 0 ≤ y ≤ 2-x ---- for region II
The required probability
= [tex]\int_{0}^{0.5} \int_{0.5-x}^{1.5}2x^3 dy dx+\int_{0.5}^{1} \int_{0}^{2-x}2x^3 dy dx[/tex]
= [tex]\int_{0}^{0.5} 2x^3[\int_{0.5-x}^{1.5}dy] dx+\int_{0.5}^{1} 2x^3[\int_{0}^{2-x}dy] dx[/tex]
= [tex]\int_{0}^{0.5} 2x^3(1.5-0.5+x) dx+\int_{0.5}^{1} 2x^3 (2-x) dx[/tex]
= [tex]\int_{0}^{0.5}2x^3 dx+\int_{0}^{0.5}2x^4 dx+\int_{0.5}^{1}4x^3 dx-\int_{0.5}^{1}2x^4[/tex]
= [tex][\frac{2x^4}{4}]_0^{0.5}+[\frac{2x^5}{5}]_0^{0.5}+[\frac{4x^4}{4}]_{0.5}^1-[\frac{2x^5}{5}]_{0.5}^1[/tex]
= [tex]$\frac{1}{32}+\frac{1}{80}+\frac{1}{16}-\frac{31}{20}$[/tex]
= 19/32
=0.59
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A technician compares repair costs for two types of microwave ovens (type I and type II). He believes that the repair cost for type I ovens is greater than the repair cost for type II ovens. A sample of 61 type I ovens has a mean repair cost of $79.60, with a standard deviation of $23.47. A sample of 45 type II ovens has a mean repair cost of $75.25, with a standard deviation of $15.07. Conduct a hypothesis test of the technician's claim at the 0.05 level of significance. Let μ1 be the true mean repair cost for type I ovens and μ2 be the true mean repair cost for type II ovens.Step 3 of 4 : Determine the decision rule for rejecting the null hypothesis H0. Round the numerical portion of your answer to three decimal places.
Answers
The question is missing some parts. Here is the complete question.
A technician compares repair costs for two types of microwave ovens (type I and type II). He believes that the repair cost for type I ovens is greater than the repair cost in type II ovens. A samplle of 61 type I ovens has a mean repair cost of $79.60, with a standard deviation of $23.47. A sample of 45 type II ovens has a mean repair cost of $75.25, with standar deviation of $15.07. Conduct a hypothesis test of the technician's claim at the 0.05 level of significance. Let [tex]\mu_{1}[/tex] be the true mean repair cost for type I ovens and [tex]\mu_{2}[/tex] be the true mean repair cost for type II ovens.
Step 1 of 4: State null and alternative hypothesis for the test.
Step 2 of 4: Compute the value of test statistics. Round your answer to 2 decimal places.
Step 3 of 4: Determine the decision rule for rejecting the null hypothesis. Round the numerical portion of your answer to 3 decimal places.
Step 4 of 4: Make the decision for the hypothesis test. (Reject or fail to reject Null Hypothesis)
Answer and Step-by-step explanation:
First, state null and alternative hypothesis:
[tex]H_{0}: \mu_{1}=\mu_{2}[/tex]
[tex]H_{a}: \mu_{1}>\mu_{2}[/tex]
Use z-statistics:
[tex]z = \frac{x_{1}-x_{2}}{y}[/tex]
where x₁ and x₂ are the means and:
[tex]y=\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}} }[/tex]
Calculating value of test stattistic:
[tex]y=\sqrt{\frac{(23.47)^{2}}{61}+\frac{75.25^{2}}{45} }[/tex]
[tex]y = \sqrt{9.030+5.046}[/tex]
[tex]y = \sqrt{14.076}[/tex]
y = 3.752
[tex]z = \frac{79.6-75.25}{3.752}[/tex]
z = 1.16
The test statistics is z = 1.16
The decision rule for rejecting null hypothesis is: [tex]\alpha[/tex] ≤ test statistics
Using z-score table, it is possible to determine p-value, which is:
p-value = 0.877
Comparing p-value with level of significance (α = 0.05):
0.877 > 0.05
Therefore, we fail to reject the null hypothesis.
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Answer:
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Step-by-step explanation:
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Answer:
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Step-by-step explanation:
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Answer:
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Step-by-step explanation:
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Answer:
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Step-by-step explanation:
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Step-by-step explanation:
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Step-by-step explanation:
Hey there!
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Answers
Answer:
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Answers
Answer:
Let m be the minimum value of original data set.
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Answer:
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Answers
Answer:
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Answer:
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Answer:
Step-by-step explanation:
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Step-by-step explanation:
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Answer:
with what?
Step-by-step explanation:
Answer:
Step-by-step explanation:
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Step-by-step explanation:
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Answer:
Inez is correct.
Step-by-step explanation:
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Answer
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Step-by-step explanation:
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Answer:
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Step-by-step explanation:
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Step-by-step explanation:
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Answer:
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Step-by-step explanation:
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Step-by-step explanation:
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Answers
Answer:
$15
Step-by-step explanation:
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Answers
Answer:
False
Step-by-step explanation:
Only six states have two syllables in their name.
No, only six states have at least two syllables.
Here,
We have to check the given statement '' All state names have at least two syllables'' is true or false.
What are states which have at least two syllables?
There are only 6 states have at least two syllables.
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Now,
There are only 6 states have at least two syllables.
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Step-by-step explanation:
do 3 2/5 times by 7 to get ur answer
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Answers
2x-1=7
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Divide both sides - x=4