Answer:
0.018 is the probability that a randomly selected adult has an IQ greater than 131.5
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 100
Standard Deviation, σ = 15
We are given that the distribution of IQ score is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
a) P(IQ greater than 131.5)
P(x > 131.5)
[tex]P( x > 131.5) = P( z > \displaystyle\frac{131.5 - 100}{15}) = P(z > 2.1)[/tex]
[tex]= 1 - P(z \leq 2.1)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x > 131.5) = 1 - 0.982 = 0.018 = 1.8\%[/tex]
0.018 is the probability that a randomly selected adult has an IQ greater than 131.5
You are given the probability that an event will not happen. Find the probability that the event will happen.
1. P(E') = 0.14
2. P(E') = 0.92
3. P(E') = 17/35
4. P(E') = 61/100
Answer:
1) P(E) = 1 - 0.14 = 0.86
2) P(E) = 1 - 0.92 = 0.08
3) P(E) = 1 - 17/35 = 18/35
4) P(E) = 1 - 61/100 = 39/100
Step-by-step explanation:
The probability that an event will happen equals one minus the probability that an event will not happen.
P(E) = 1 - P(E') ......1
Where;
P(E) is the probability that an event will happen.
P(E') is the probability that an event will not happen.
1. P(E') = 0.14
Applying equation 1
P(E) = 1 - 0.14 = 0.86
2. P(E') = 0.92
P(E) = 1 - 0.92 = 0.08
3. P(E') = 17/35
P(E) = 1 - 17/35 = 18/35
4. P(E') = 61/100
P(E) = 1 - 61/100 = 39/100
Farmer Bob's dairy farm is ready for milking day. His 199 cows line up in the barn and he begins milking them. The first cow gives 4 pints of milk. The next cow gives 5 pints of milk. The next gives 6 pints, and so on, increasing by 1 pint each cow. How many pints of milk does the last cow give?
the last cow gives 202 pints of milk
Given :
There are 199 cows line up in the barn and he begins milking them.
The first cow gives 4 pints of milk. The next cow gives 5 pints of milk. The next gives 6 pints, and so on, increasing by 1 pint each cow.
pints of milk is increasing 1 by a sequence
4,5,6,7 ......... and so on
when n=1 , the cow given 4 pints of milk
The sequence is arithmetic
nth cow is 199. Lets find out pints of milk when n=199
apply nth formula of arithmetic
[tex]T_n = T_1 + (n-1) d\\[/tex]
n=199
difference d=1
T1=4
Substitute all the values
[tex]T_n = T_1 + (n-1) d\\\\\\T_{199}=4+(199-1)1\\T_{199}=4+198\\T_{199}=202[/tex]
So, the last cow gives 202 pints of milk
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Five cards are drawn from an ordinary deck of 52 playing cards. What is the probability that the hand drawn is a full house? (A full house is a hand that consists of two of one kind and three of another kind.)
Answer:
The required probability is 0.00144 or 0.144%.
Step-by-step explanation:
Consider the provided information.
Five cards are drawn from an ordinary deck of 52 playing cards.
A full house is a hand that consists of two of one kind and three of another kind.
The total number of ways to draw 5 cards are: [tex]^{52}C_5=\frac{52!}{5!47!}[/tex]
Now we want two of one kind and three of another.
Let the hand has the pattern AAABB, where A and B are from distinct kinds. The number of such hands are:
[tex]^{13}C_1\times^{4}C_3\times^{12}C_1\times^{4}C_2=\frac{13!}{12!}\times\frac{4!}{3!}\times\frac{12!}{11!}\times\frac{4!}{2!2!}[/tex]
Thus, the required probability is:
[tex]\frac{^{13}C_1\times^{4}C_3\times^{12}C_1\times^{4}C_2}{^{52}C_5}=\frac{\frac{13!}{12!}\times\frac{4!}{3!}\times\frac{12!}{11!}\times\frac{4!}{2!2!}}{\frac{52!}{5!47!}}[/tex]
[tex]=\frac{3744}{2598960}\\\\\approx0.00144[/tex]
Hence, the required probability is 0.00144 or 0.144%.
The sample space of a random experiment is {a,b,c,d,e} with probabilities 0.1,0.1,0.2,0.4, and 0.2, respectively. Let A denote the event {a,b,c}, and let B denote the even t {c,d,e}. Determine the following:
a. P(A)
b. P(B)
c. P(A’)
d. P(AUB)
e. P(AnB)
Answer with Step-by-step explanation:
We are given that a sample space
S={a,b,c,d,e}
P(a)=0.1
P(b)=0.1
P(c)=0.2
P(d)=0.4
P(e)=0.2
a.A={a,b,c}
P(A)=P(a)+P(b)+P(c)
P(A)=0.1+0.1+0.2=0.4
b.B={c,d,e}
P(B)=P(c)+P(d)+P(e)=0.2+0.4+0.2=0.8
c.A'=Sample space-A={a,b,c,d,e}-{a,b,c}={d,e}
P(A')=P(d)+P(e)=0.4+0.2=0.6
d.[tex]A\cup B[/tex]={a,b,c,d,e}
[tex]P(A\cup B)[/tex]=P(a)+P(b)+P(c)+P(d)+P(e)=0.1+0.1+0.2+0.4+0.2=1
e.[tex]A\cap B[/tex]={c}
[tex]P(A\cap B)=P(c)=0.2[/tex]
A roofing contractor purchases a shingle delivery truck with a shingle elevator for $42,000. The vehicle requires an average expenditure of $9.50 per hour for fuel and maintenance, and the operator is paid $11.50 per hour. Write a linear equation giving the total cost C of operating this equipment for t hours. (Include the purchase cost of the equipment.)
Answer:
[tex]T(t) = 42000 + 21t[/tex]
Step-by-step explanation:
There are two costs for the operation of the equipment, the purchase cost and the hourly cost. The total cost is the sum of these two costs:
So
[tex]T = F_{c} + H_{c}[/tex]
In which [tex]F_{c}[/tex] is the fixed cost and [tex]H_{c}[/tex] is the hourly cost.
Fixed cost
A roofing contractor purchases a shingle delivery truck with a shingle elevator for $42,000. This means that the fixed cost is 42000. So [tex]F_{c} = 42000[/tex]
Hourly cost
The vehicle requires an average expenditure of $9.50 per hour for fuel and maintenance, and the operator is paid $11.50 per hour. So for each hour, there are expenses of 9.5 + 11.5 = $21.
So the hour cost is
[tex]H_{c}(t) = 21t[/tex]
In which t is the number of hours
Total cost
[tex]T = F_{c} + H_{c}[/tex]
[tex]T(t) = 42000 + 21t[/tex]
Rewrite each of the following statements in the form "∀ _____ x, _____." (a) All dinosaurs are extinct. ∀ x, . (b) Every real number is positive, negative, or zero. ∀ x, . (c) No irrational numbers are integers. ∀ x, .(d) No logicians are lazy. ∀ x, .(e) The number 2,147,581,953 is not equal to the square of any integer. ∀ x, .(f) The number −1 is not equal to the square of any real number.
Answer:
See below
Step-by-step explanation:
Essentially, we have to replace "quantifier words" like "All", "Every" by the universal quantifier ∀.
a) ∀ dinosaur x, x is extinct.
b) ∀ real number x, x is positive, negative, or zero.
c) ∀ irrational number x, x is not an integer.
d) ∀ logician x, x is not lazy.
e) ∀ integer x, x²≠ 2,147,581,953.
f) ∀ real number x, x²≠ -1.
In a) and b) we replace the words without major changes. In the other statements, we modify the statement using negation. For example, "No irrational numbers are integers." is equivalent to "Every irrational number is not integer".
Western Athletic Club International (WACI) owns and operates a chain of fitness clubs. Currently WACI has 675 members at its Collegetown location, a 13% increase over the previous year. Next year, WACI hopes to grow by the same number of members. 66% of its members are women. What percent increase will next year's growth represent?
Answer:
12%
Step-by-step explanation:
let x represent the actual number of members last year.
current members = 675 and with an increment of 13% over previous year
to find the increment
(675 - x) / x = 0.13
675 - x = 0.13x
collect the like terms
675 = 0.13x + x
675 = 1.13 x
x = approx 597 members were there last year
its hope to increase by the same number next year
percent increase = 675 - 597 / 675 = 78 / 675 = 0.12 = 12%
At what points does the helix r(t) = sin(t), cos(t), t intersect the sphere x2 + y2 + z2 = 17? (Round your answers to three decimal places. If an answer does not exist, enter DNE.)
Answer:
the helix intersects the sphere at t=4 and t=(-4)
Step-by-step explanation:
for the helix r(t) = [ sin(t) , cos(t) , t ] then x=sin(t) , y=cos(t) and z=t
thus the helix intersect the sphere x² + y² + z² = 17 at
x² + y² + z² = 17
[sin(t)]²+[cos(t)]²+ t² = 17
1 + t² = 17
t² = 16
t = ±4
thus the helix intersects the sphere at t=4 and t=(-4)
The point wher the helix intersect the sphere is at t = ±4
The coordinate of a helixGiven the coordinate of a helix expressed as;
r(t) = [sin(t), cos(t), t]If these coordinate intersects the sphere x² + y² + z² = 17
Substitute the coordinate of the helix:
x² + y² + z² = 17
(sint)² + (cost)² + t² = 17
sin²t + cos²t + t² = 17
1 + t² = 17
t² = 17 - 1
t² = 16
t = ±√16
t = ±4
Hence the point wher the helix intersect the sphere is at t = ±4
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Suppose your statistics instructor gave six examinations during the semester. You received the following grades (percent correct): 79, 64, 84, 82,92, and 77. Instead of averaging the six scores, the instructor indicated he would randomly select two grades and report that grade to the student records office.
a. How many different samples of two test grades are possible?
b. List all possible samples of size two and compute the mean of each.
c. Compute the mean of the sample means and compare it to the population mean.
Answer:
a. 15
b.
Sr.no Samples Sample mean
1 (79,64) 71.5
2 (79,84) 81.5
3 (79,82) 80.5
4 (79,92) 85.5
5 (79,77) 78
6 (64,84) 74
7 (64,82) 73
8 (64,92) 78
9 (64,77) 70.5
10 (84,82) 83
11 (84,92) 88
12 (84,77) 80.5
13 (82,92) 87
14 (82,77) 79.5
15 (92,77) 84.5
c.
mean of sample mean=population mean=79.67
Step-by-step explanation:
a.
The different samples of two test grade are nCr, where n=6 and r=2.
nCr=6C2=6!/2!(6-2)!=6*5*4!/2!4!=30/2=15.
Thus, there are 15 different samples of two test grade.
b.
All possible samples are listed below:
Sr.no Samples
1 (79,64)
2 (79,84)
3 (79,82)
4 (79,92)
5 (79,77)
6 (64,84)
7 (64,82)
8 (64,92)
9 (64,77)
10 (84,82)
11 (84,92)
12 (84,77)
13 (82,92)
14 (82,77)
15 (92,77)
The sample means for each sample can be calculated as
Sr.no Samples Sample mean
1 (79,64) (79+64)/2=71.5
2 (79,84) (79+84)/2=81.5
3 (79,82) (79+82)/2=80.5
4 (79,92) (79+92)/2=85.5
5 (79,77) (79+77)/2=78
6 (64,84) (64+84)/2=74
7 (64,82) (64+82)/2=73
8 (64,92) (64+92)/2=78
9 (64,77) (64+77)/2=70.5
10 (84,82) (84+82)/2=83
11 (84,92) (84+92)/2=88
12 (84,77) (84+77)/2=80.5
13 (82,92) (82+92)/2=87
14 (82,77) (82+77)/2=79.5
15 (92,77) (92+77)/2=84.5
c.
The sample means of sample mean μxbar will calculated by taking average of sample means
μxbar=(71.5+ 81.5+ 80.5+ 85.5+ 78+ 74+ 73+ 78+ 70.5+ 83+ 88+ 80.5+ 87+ 79.5+ 84.5)/15
μxbar=1195/15=79.67
Population mean=μ=(79+64+84+82+92+77)/6
μ=478/6=79.67
Sample means of sample mean μxbar=Population mean μ.
Suppose that prices of a gallon of milk at various stores in one town have a mean of $3.94 with a standard deviation of $0.11. Using Chebyshev's Theorem, state the range in which at least 88.9% of the data will reside. Please do not round your answers.
Answer:
($3.61, $4.27) is the range in which at least 88.9% of the data will reside.
Step-by-step explanation:
Chebyshev's Theorem
This theorem states that at least [tex]1 - \dfrac{1}{k^2}[/tex] percentage of data lies within k standard deviation from the mean.For k = 2[tex]\mu \pm 2\sigma\\\\1 - \dfrac{1}{(2)^2}\% = 75\%[/tex]
Atleast 75% of data lies within two standard deviation of the mean for a non-normal data.
For k = 3[tex]\mu \pm 3\sigma\\\\1 - \dfrac{1}{(3)^2}\% = 88.9\%[/tex]
Thus, range of data for which 88.9% of data will reside is
[tex]\mu \pm 3\sigma\\= (\mu - 3\sigma , \mu + \3\sigma)[/tex]
Now,
Mean = $3.94
Standard Deviation = $0.11
Putting the values, we get,
Range =
[tex]3.94 \pm 30.11\\= (3.94 - 3(0.11) , 3.94 + 3(0.11))\\=(3.61,4.27)[/tex]
($3.61, $4.27) is the range in which at least 88.9% of the data will reside.
Answer:
Using Chebyshev's Theorem, the range in which at least 88.9% of the data reside is [$3.82, $4.06].
Step-by-step explanation:
Given:
The mean of the prices of a gallon of milk is, [tex]\mu=\$3.94[/tex]
The standard deviation of the prices of a gallon of milk is, [tex]\sigma=\$0.11[/tex]
The Chebyshev's Theorem states that for a random variable X with finite mean μ and finite standard deviation σ and for a positive constant k the provided inequality exists,
[tex]P(|X-\mu|\geq k)\leq \frac{\sigma^{2}}{k^{2}}[/tex]
The value of [tex]\frac{\sigma^{2}}{k^{2}} = 0.889[/tex]
Then solve for k as follows:
[tex]\frac{\sigma^{2}}{k^{2}} = 0.889\\\frac{(0.11)^{2}}{k^{2}}=0.889\\k^{2}=\frac{(0.11)^{2}}{0.889}\\k=\sqrt{0.0136108} \\\approx0.1167[/tex]
The range in which at least 88.9% of the data will reside is:
[tex]P(|X-\mu|\geq k)\leq \frac{\sigma^{2}}{k^{2}}\\P(\mu-k\leq X\leq \mu+k)\leq 0.889\\P(3.94-0.1167\leq \leq X\leq 3.94+0.1167)\leq 0.889\\P(3.8233\leq X\leq 4.0567)\leq 0.889\\\approxP(3.82\leq X\leq 4.06)\leq 0.889[/tex]
Thus, the probability of prices of a gallon of milk between $3.82 and $4.06 is 0.889.
Prior Knowledge Questions (Do these BEFORE using the Gizmo.) Imagine you and your friends are making hot dogs. A complete hot dog consists of a wiener and a bun. At the store, you buy four packages of eight wieners and three bags of 10 buns. 1. How many total hot dogs can you make? ________________________________________ 2. Which ingredient limited the number of hot dogs you could make? ____________________ 3. Which ingredient will you have leftovers of? ______________________________________
Answer:
A. 8 complete hotdogs were made.
B. The wiener.
C. The bun.
Step-by-step explanation:
1 complete hotdog = 1 wiener + 1 bun
You have, 3*10 bun and 8 wiener
= 30 bun and 8 wiener
A. Since, we have 30 bun and 8 wiener and for 1 complete hotdog = 1 wiener + 1 bun
Therefore, 8 hotdogs = 8 wiener + 8 buns
= 8 complete hotdogs are made
B.
Since 8 complete hotdogs are made, 8 wiener and 8 buns are used.
Therefore, the limiting ingredient is the wiener because it is the smallest number of ingredient and none of it was left after making the 8 hotdogs.
C. Since 8 complete hotdogs are made, 8 buns were used
So, the leftover buns = (30 - 8) buns
= 22 buns
The buns is the leftover ingredient
At the store, you buy four packages of eight wieners and three bags of 10 buns, you can make 8 hotdogs.
There are 8 hotdogs you can make then the limited ingredient is wieners.
The buns are the leftover ingredient.
Given that,You and your friends are making hot dogs.
A complete hot dog consists of a wiener and a bun.
At the store, you buy four packages of eight wieners and three bags of 10 buns.
According to the question,Total number of buns = 30
Total number of wieners = 8
At the store, you buy four packages of eight wieners and three bags of 10 buns.
For every hot dog, 1 wiener one 1 buns are required. the calculation for the given question is given as follows;
1. The total number of hot dogs can you make is,
To make 1 hot dog 1 wiener and 1 bun
Then,
for 8 hot dogs, 8 wieners and 8 buns are required.
Therefore, you can make 8 hotdogs.
2. There are 8 hotdogs you can make then the limited ingredient is wieners.
3. After making 8 hot dogs only 8 buns are used,
Then,
Left buns = Total number of buns - used buns
Left buns = 30-8 = 22
The buns are the leftover ingredient.
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"Consider the following argument:
a. George and Mary are not both innocent.
b. If George is not lying, Mary must be innocent.
c. Therefore, if George is innocent, then he is lying.
Let ???? be the proposition "George is innocent", m be the proposition "Mary is innocent", and let ???? be the proposition "George is lying"."
1. Write a propositional formula F involving variables g, m, l such that the above argument is valid if and only if F is valid.
2. Is the above argument valid? If so, prove its validity by proving the validity of F. If not, give an interpretation under which F evaluates to false.
College Mathematics 10+5 pts
"Consider the following argument:
a. George and Mary are not both innocent.
b. If George is not lying, Mary must be innocent.
c. Therefore, if George is innocent, then he is lying.
Let ???? be the proposition "George is innocent", m be the proposition "Mary is innocent", and let ???? be the proposition "George is lying"."
1. Write a propositional formula F involving variables g, m, l such that the above argument is valid if and only if F is valid.
2. Is the above argument valid? If so, prove its validity by proving the validity of F. If not, give an interpretation under which F evaluates to false.
Answer
1. [tex]F = ((g\cap \sim m)\cup(\sim g\cap m)\cap(\sim l\to m))\to(g\to l) [/tex]
2. It is valid. See explanation for proof.
Step-by-step explanation
1. Statement a is an exclusive OR of [tex]g[/tex] and [tex]m[/tex]. This is because only one of them is innocent while the other is not. That is [tex]g[/tex] is true and [tex]m[/tex] is false or [tex]g[/tex] is false and [tex]m[/tex] is true.
Statement b is an implication that [tex]\sim l \to m[/tex].
Statement c is another implication that [tex]g\to l[/tex].
The presence of the word "therefore" in statement c means it is an implication from statement a AND statement b. So we have
[tex]F = ((g\cap \sim m)\cup(\sim g\cap m)\cap(\sim l\to m))\to(g\to l) [/tex]
2.
The validity will be proved using a truth table. [tex]F[/tex] is valid if the last column contains only true values i.e. truth values of T.
From the table (in the attached image), [tex]c[/tex] represents the statement a, [tex]d[/tex] represents statement b and [tex]f[/tex] represents statement c. The last column, which represents [tex]F[/tex], is valid as all its entries are T.
To translate the argument into logic notation, one could write it as ((¬g ∨ m) ∧ (¬l → m)) → ((g → ¬l). Upon considering the case where both George and Mary are innocent, and George is not lying, it can be observed that the argument is invalid.
Explanation:1. The propositional formula F can be written as: ((¬g ∨ m) ∧ (¬l → m)) → ((g → ¬l). This translates the given argument into logical notation.
2. This argument is not valid. An easy way to prove this is by considering a scenario where both George and Mary are innocent, hence g and m are both true, and let l be the proposition 'George is lying', so l is false. In this case, the left-hand side of your main implication results true ((¬g ∨ m) ∧ (¬l → m)) = (false ∨ true) ∧ (true → true) = true ∧ true = true), but the right-hand side of your main implication results false (g → ¬l = true → false = false) hence the full implication results false, which makes the argument invalid.
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A forensic psychologist studying the accuracy of a new type of polygraph (lie detector) test instructed a participant ahead of time to lie about some of the questions asked by the polygraph operator. On average, the current polygraph test is 75% accurate, with a standard deviation of 6.5%. With the new machine, the operator correctly identified 83.5% of the false responses for one participant. Using the.05 level of significance, is the accuracy of the new polygraph different from the current one? Fill in the following information: Assuming an ?-0.05, determine the z-score cutoff for the rejection region. Calculate the test statistic for the given data Zobt Based on the data above, finish the statement about your decision: Based on the observed z-score, we would decide to (accept, reject, fail to reject, fail to accept) hypothesis. the (null, alternative)
Answer:
[tex]z=\frac{83.5-75}{6.5}=1.31[/tex]
The rejection zone for this case would be:
[tex] z> 1.96 \cup Z<-1.96[/tex]
[tex]p_v =2*P(z>1.31)=0.1901[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is significantly different from 75 at 5% of significance.
Step-by-step explanation:
Data given and notation
[tex]\bar X=83.5[/tex] represent the sample mean
[tex]\sigma=6.5[/tex] represent the population standard deviation for the sample
[tex]\mu_o =75[/tex] represent the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is different from 75, the system of hypothesis would be:
Null hypothesis:[tex]\mu = 75[/tex]
Alternative hypothesis:[tex]\mu \neq 75[/tex]
For this case we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:
[tex]z=\frac{\bar X-\mu_o}{\sigma}[/tex] (1)
z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]z=\frac{83.5-75}{6.5}=1.31[/tex]
Cutoff for the rejection regon
Since our significance level is [tex] \alpha=0.05[/tex] and we are conducting a bilateral test we need to find a quantile in the standard normal distribution that accumulates 0.025 of the area on each tail.
And for this case those values are [tex] z_{crit}= \pm 1.96[/tex]
So the rejection zone for this case would be:
[tex] z> 1.96 \cup Z<-1.96[/tex]
Our calculated value is not on the rejection zone. So we fail to reject the null hypothesis.
P-value
Since is a two sided test the p value would be given by:
[tex]p_v =2*P(z>1.31)=0.1901[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is significantly different from 75 at 5% of significance.
Final answer:
To determine if the new polygraph test's accuracy significantly differs from the current one, we compare the calculated test statistic with the critical z-score of ±1.96 at an alpha level of 0.05. The process involves hypothesis testing, where rejecting or failing to reject the null hypothesis depends on whether the test statistic exceeds the critical value.
Explanation:
The question is about determining if the accuracy of a new type of polygraph test is statistically different from the current one using hypothesis testing. Given an alpha level (α) of 0.05 (5% level of significance), the critical z-score for a two-tailed test is ±1.96. This critical value defines the cutoff points for the rejection regions. To calculate the test statistic (Zobt), we use the formula:
Zobt = (X - μ) / (σ / √n),
where X is the sample mean (83.5%), μ is the population mean (75%), σ is the standard deviation (6.5%), and √n is the square root of the sample size. Given that only one participant's data is used, √n is 1 for this scenario, which simplifies our formula. Unfortunately, without the exact sample size for a more precise calculation or this being a theoretical scenario with one participant, we remark on the process rather than providing a numerical value for Zobt.
Based on the calculated Zobt, if it exceeds the critical z-score (1.96 or -1.96), we reject the null hypothesis indicating that there is a statistically significant difference between the new polygraph's accuracy and the current one. If Zobt does not exceed the critical value, we fail to reject the null hypothesis, indicating no significant difference.
An air traffic controller has noted that it clears an average of seven planes per hour for landing. What is the probability that during the next two hours exactly 15 planes will be cleared for landing?a. 0.0989 b. Not enough information is given to answer the problem. c. 0.0033 d. 0.0651
Answer:
a) 0.0989, Option a
Step-by-step explanation:
The concept of Poisson probability distribution is used as shown in the attached file.
A space vehicle has an independent backup system for one of its communication networks. The probability that either system will function satisfactorily during a flight is 0.985. What is the probability that during a given flight (a) both systems function satisfactorily, (b) at least one system functions satisfactorily, and (c) both systems fail?
Answer:
(a) 0.970225.
(b) 0.999775.
(c) 2.25 x [tex]10^{-4}[/tex].
Step-by-step explanation:
Given the probability that either system will function satisfactorily during a flight is 0.985.
(a) To calculate the probability that both systems function satisfactorily we are given the probability of either system will function of 0.985 which means that both function have probability of functioning satisfactorily of 0.985 each i.e.,
= 0.985 x 0.985 = 0.970225.
(b) The probability of first function will not function satisfactorily is 1 - 0.985 because the probability of first system function satisfactorily is 0.985.
Similarly, probability of second function will not function satisfactorily is also 1 - 0.985 = 0.015
So Probability that during a given flight at least one system functions satisfactorily = 1 - both system will not function satisfactorily
= 1 - (0.015 x 0.015) = 0.999775.
(c) Probability that during a given flight both systems fail or not function satisfactorily = (1 - 0.985) x (1 - 0.985) = 2.25 x [tex]10^{-4}[/tex]
Answer:
(a) P (Both systems functioning satisfactorily) = 0.970
(b) P (At least one system functions satisfactorily) = 0.999
(c) P (Both the systems failing) = 0.00023
Step-by-step explanation:
Let A = the system in use is working satisfactorily and B = the backup system is working satisfactorily.
The probability that either of the systems works satisfactorily is,
P (A) = P(B) = 0.985
(a)
Both the events A and B are independent, i.e. [tex]P(A\cap B)=P(A)\times P(B)[/tex]
Compute the probability that during a given flight both systems function satisfactorily as follows:
[tex]P(A\cap B)=P(A)\times P(B)[/tex]
[tex]=0.985\times0.985\\=0.970225\\\approx0.970[/tex]
Thus, the probability of both systems functioning satisfactorily is 0.970.
(b)
Compute the probability that during a given flight at least one system functions satisfactorily as follows:
P (At least one system functions satisfactorily) = 1 - P (None of the system functions satisfactorily)
= [tex]1-[P(A^{c})\times P(B^{c})][/tex]
[tex]=1-([1-P(A)]\times [1-P(B)])\\=1-([1-0.985]\times [1-0.985])\\= 1-0.000225\\=0.999775\\\approx0.999[/tex]
Thus, the probability that during a given flight at least one system functions satisfactorily is 0.999.
(c)
Compute the probability that during a given flight both the systems fail as follows:
P (Both the systems failing) = [tex]P(A^{c})\times P(B^{c})[/tex]
[tex]=[1-P(A^{c})]\times [1-P(B^{c})]\\=(1-0.985)\times (1-0.985)\\=0.015\times 0.015\\=0.000225\\\approx0.00023[/tex]
Thus, the probability that during a given flight both the systems fail is 0.00023.
Suppose you draw a card from a well shuffled deck of 52 what is the probability of drawing a 10 or jack
The probability of drawing a 10 or Jack will be 8/52.
What is probability?The chances of an event occurring are defined by probability. Probability has several uses in games, and in business to create probability-based forecasts.
The number of jack cards and 10 number cards from the pack of cards is 4,
The probability of drawing a 10 or jack is,
P = P(10) + P(J)
P = (4/52)+(4/52)
P= 8/52
Hence, the probability of drawing a 10 or Jack will be 8/52.
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The average time entities spend in the system in five simulation runs are: 25.2, 19.7, 23.6, 18.6, and 21.4 minutes, respectively. Five more simulations are run and the following average times in the system are obtained 22.1, 26.0, 20.2, 16.4, and 17.9 minutes. a). Build a 95% confidence interval for the mean time in the system using the first five averages collected.
Answer:
a) [tex]21.7-2.776\frac{2.718}{\sqrt{5}}=18.33[/tex]
[tex]21.7+2.776\frac{2.718}{\sqrt{5}}=25.07[/tex]
So on this case the 95% confidence interval would be given by (18.33;25.07)
b) [tex]20.52-2.776\frac{3.757}{\sqrt{5}}=18.84[/tex]
[tex]20.52+2.776\frac{3.757}{\sqrt{5}}=22.20[/tex]
So on this case the 95% confidence interval would be given by (18.84;22.20)
c) [tex]21.11-2.262\frac{3.154}{\sqrt{10}}=18.85[/tex]
[tex]21.11+2.262\frac{3.154}{\sqrt{10}}=23.37[/tex]
So on this case the 95% confidence interval would be given by (18.85;23.37)
And as we can see the confidence intervals are very similar for the 3 cases.
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
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".
[tex]\bar X[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s represent the sample standard deviation
n represent the sample size
Part a) Build a 95% confidence interval for the mean time in the system using the first five averages collected.
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
Data: 25.2, 19.7, 23.6, 18.6, and 21.4
In order to calculate the mean and the sample deviation we can use the following formulas:
[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)
[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)
The mean calculated for this case is [tex]\bar X=21.7[/tex]
The sample deviation calculated [tex]s=2.718[/tex]
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=5-1=4[/tex]
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,4)".And we see that [tex]t_{\alpha/2}=2.776[/tex]
Now we have everything in order to replace into formula (1):
[tex]21.7-2.776\frac{2.718}{\sqrt{5}}=18.33[/tex]
[tex]21.7+2.776\frac{2.718}{\sqrt{5}}=25.07[/tex]
So on this case the 95% confidence interval would be given by (18.33;25.07)
Part b: Build a 95% confidence interval for the mean time in the system using the second set of five averages collected.
Data: 22.1, 26.0, 20.2, 16.4, and 17.9
The mean calculated for this case is [tex]\bar X=20.52[/tex]
The sample deviation calculated [tex]s=3.757[/tex]
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=5-1=4[/tex]
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,4)".And we see that [tex]t_{\alpha/2}=2.776[/tex]
Now we have everything in order to replace into formula (1):
[tex]20.52-2.776\frac{3.757}{\sqrt{5}}=18.84[/tex]
[tex]20.52+2.776\frac{3.757}{\sqrt{5}}=22.20[/tex]
So on this case the 95% confidence interval would be given by (18.84;22.20)
Part c: Build a 95% confidence interval for the mean time in the system using all ten averages collected.
Data: 25.2, 19.7, 23.6, 18.6, 21.4, 22.1, 26.0, 20.2, 16.4, and 17.9
The mean calculated for this case is [tex]\bar X=21.11[/tex]
The sample deviation calculated [tex]s=3.154[/tex]
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=10-1=9[/tex]
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,9)".And we see that [tex]t_{\alpha/2}=2.262[/tex]
Now we have everything in order to replace into formula (1):
[tex]21.11-2.262\frac{3.154}{\sqrt{10}}=18.85[/tex]
[tex]21.11+2.262\frac{3.154}{\sqrt{10}}=23.37[/tex]
So on this case the 95% confidence interval would be given by (18.85;23.37)
And as we can see the confidence intervals are very similar for the 3 cases.
How many ways are there to pick a group of n people from 100 people (each of a different height) and then pick a second group of m other people such that all people in the first group are taller than the people in the second group
The total number of ways to choose the two groups, depending on the relationship between n and m, is:
M < N: C(100, n).
M ≥ N: Σ (C(100, k) * C(100 - k, m - k)) for k = n to 1.
Case 1: m < n (n tallest chosen first):
Choose the n tallest people from the 100. There are C(100, n) ways to do this.
The remaining 100 - n people are all shorter than the chosen n. Since m < n, we can simply choose all the remaining people (100 - n).
The total number of ways in this case is C(100, n).
Case 2: m ≥ n (not all n tallest need to be chosen):
This case requires considering all possible combinations of how many tall people to choose (from n down to 1).
For each value of k (n, n - 1, ..., 1), choose k tallest people from the 100. There are C(100, k) ways to do this.
The remaining 100 - k people are all shorter than the chosen k. Choose the remaining m - k people from this group. There are C(100 - k, m - k) ways to do this.
Sum the results for each value of k: Σ (C(100, k) * C(100 - k, m - k)) for k = n to 1.
Therefore, the total number of ways to choose the two groups, depending on the relationship between n and m, is:
M < N: C(100, n)
M ≥ N: Σ (C(100, k) * C(100 - k, m - k)) for k = n to 1
Complete question:
How many ways are there to pick a group of n people from 100 people, and then pick a second group of m people from the remaining 100 − n people, so that all the people you chose from the first group are taller than the people from the second group (assume that everyone in the group of 100 people has a different height.)
Note: it is implied in this question that 1 ≤ n ≤ 100 and 0 ≤ m ≤ 100 − n.
kevin durant of the oklahoma city thunder & kobe bryant of the los angeles lakers were the leading scorers in the NBA for the 2012-2013 regular season. Together they scored 4413 points, with bryant scoring 147 fewer points than durant. how many pointsdid each of them score?
Answer:
Bryant: 2059.5 (RIP)
Durant:2353.5
Step-by-step explanation:
To find how many points Kevin Durant and Kobe Bryant scored, set up equations based on the given information. Kevin Durant scored 2280 points and Kobe Bryant scored 2133 points during the 2012-2013 NBA regular season.
Explanation:The question asks us to determine how many points Kevin Durant and Kobe Bryant scored individually during the 2012-2013 NBA regular season given that together they scored 4413 points and Kobe Bryant scored 147 fewer points than Kevin Durant.
To solve this, we can set up two equations based on the information provided:
We can substitute the second equation into the first to find Durant's score:
Now that we have Durant's score, we can use it to find Bryant's score:
Therefore, Kevin Durant scored 2280 points and Kobe Bryant scored 2133 points during the 2012-2013 NBA regular season.
Bottled water and medical supplies are to be shipped to victims of a hurricane by plane. Each plane can carry 90,000 pounds and a total volume of 6000 cubic feet. The bottled water weighs 20 pounds per container and measures 1 cubic foot. The medical kits each weigh 10 pounds and measure 2 cubic feet.
(a) How many containers of bottled water and how many medical kits can be sent on each plane?
Answer:4000 bottle containers and 1000
Medical kits
Step-by-step explanation:
The total weights the plane can take per trip is 90000lb. It implies that if we multiply the weight of each bottle container by the number of bottle container and add it to the product of the number medical kit and the weight of each medical kit, we will obtain a total weight of 90000lb.
If x is the number of bottle and y is the number of medical kit
20×x+10×y=90000....i
Also
The total volume the plane can take per trip is 6000ft³ It implies that if we multiply the volume of each bottle container by the number of bottle container and add it to the product of the number medical kit and the volume of each medical kit, we will obtain a total volume of 6000ft³.
Recall, x is the number of bottle and y is the number of medical kit
1×x+2×y=6000....ii
Combining equation i and ii and solving simultaneously,
x=4000 units and y= 1000 units in each plane trip
Answer:
Bottled water = 4000.
Medical kits = 1000.
Step-by-step explanation:
Let x = bottled water
y = medical kits
For the mass (in pounds):
20x + 10y = 90000
For the cubic ft:
x + 2y = 6000
Solving equation i and ii simultaneously,
x = 4000.
y = 1000.
Bottled water = 4000.
Medical kits = 1000.
Steel rods are manufactured with a mean length of 29 centimeter(cm). Because of variability in the manufacturing process, the lengths of the rods are approximately normally distributed with a standard deviation of 0.07 cm. Complete parts (a) to (d)
(a) What proportion of rods has a length less than 28.9 cm?
(b) Any rods that are shorter than 28.84 cm or longer than 29.16 cm are discarded. What proportion of rods will be discarded?
(c) Using the results of part (b) if 5000 rods are manufactured in a day, how many should the plant manager expect to discard?
(d) If an order comes in for 10,000 steel rods, how many rods should the plant manager expect to manufacture if the order states that all rods must be between
28.9 cm and 29.1 cm?
The problem involves computing z-scores and finding corresponding proportions in a normal distribution, then using these proportions to estimate rod discard rates and the amount of rods to manufacture. The mean and standard deviation of rod lengths provided are used for z-score calculations.
Explanation:In this problem, we are dealing with a normal distribution scenario where the mean (μ) is given as 29 cm and the standard deviation (σ) is 0.07 cm.
(a) Proportion of rods less than 28.9 cmTo find the proportion, we can firstly calculate the z-score using the formula z = (x - μ)/σ, where x is the required value (28.9 cm). The z-score helps us express how far our data point is from the mean in terms of standard deviations. Using standard normal distribution tables or a calculator, we can then determine the proportion of rods with lengths less than 28.9 cm.
(b) Proportion of rods to be discardedA similar strategy is used. Calculate the z-scores for 28.84 cm and 29.16 cm, then use these to find proportions. The proportion of discarded rods would be the sum of these two.
(c) Number of rods to be discardedUsing the proportion obtained in (b), multiply it by the total number of rods manufactured in a day (i.e., 5000) to get the expected number of rods to be discarded.
(d) Number of rods to be manufacturedIn this part, calculate the z-scores for 28.9 and 29.1 cm, then find the proportion of rods within these lengths using the tables or calculator. As this proportion represents the acceptable rods, divide the required amount of rods (i.e., 10,000) by this proportion to calculate the expected total rods to be manufactured to fulfill the order.
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To find the proportion of rods that have a length less than 28.9 cm, we use the standard normal distribution table. The proportion of rods that will be discarded is found by calculating the z-scores for the lower and upper bounds of the range. The expected number of discarded rods can be estimated by multiplying the proportion of discarded rods by the number of rods manufactured. The number of rods that should be manufactured within a specified range can be calculated using the proportion of rods that fall within that range.
Explanation:To solve this problem, we will use the standard normal distribution table to find the proportion of rods that have a length less than 28.9 cm. We will also use the standard normal distribution table to find the proportion of rods that will be discarded when their length is outside the range of 28.84 cm to 29.16 cm. Finally, we will use the proportion of discarded rods to estimate the number of rods that the plant manager should expect to discard when manufacturing 5000 rods in a day. For part (d), we will assume that the lengths of the rods are normally distributed and calculate the number of rods that fall within the range of 28.9 cm to 29.1 cm when manufacturing 10,000 rods.
(a) To find the proportion of rods with a length less than 28.9 cm, we need to find the z-score corresponding to 28.9 cm. The z-score formula is given by z = (x - mean) / standard deviation. Plugging in the values, we get z = (28.9 - 29) / 0.07 = -1.43. Using the standard normal distribution table, we find that the proportion of rods with a length less than 28.9 cm is 0.0764 or 7.64%.
(b) To find the proportion of rods that will be discarded, we need to find the z-scores corresponding to 28.84 cm and 29.16 cm. The z-score for 28.84 cm is z = (28.84 - 29) / 0.07 = -2.29 and the z-score for 29.16 cm is z = (29.16 - 29) / 0.07 = 2.29. Using the standard normal distribution table, we find that the proportion of rods with a length shorter than 28.84 cm or longer than 29.16 cm is 0.0485 or 4.85%. Therefore, the proportion of rods that will be discarded is 4.85%.
(c) To estimate the number of rods that the plant manager should expect to discard when manufacturing 5000 rods in a day, we multiply the proportion of discarded rods by the number of rods manufactured. The expected number of discarded rods is 0.0485 x 5000 = 242.5 or approximately 243 rods.
(d) To calculate the number of rods that the plant manager should expect to manufacture when the order states that all rods must be between 28.9 cm and 29.1 cm, we need to find the proportion of rods that fall within this range. The z-scores for 28.9 cm and 29.1 cm are calculated as z1 = (28.9 - 29) / 0.07 = -1.43 and z2 = (29.1 - 29) / 0.07 = 1.43. Using the standard normal distribution table, we find that the proportion of rods that fall within this range is 0.8471 or 84.71%. Therefore, the number of rods that the plant manager should expect to manufacture when the order states that all rods must be between 28.9 cm and 29.1 cm is 0.8471 x 10000 = 8471 rods.
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A study was done to compare people's religion with how many days that they have said they have been calm in the past three days. 0 days 1 day 2 days 3 days Protestant 50 27 6 16 Catholic 55 19 2 11 Jewish 84 34 4 18 Other 84 34 2 22 What proportion of Protestants were calm for 2 out of the 3 days? NOTE: Read the question very carefully.
Using it's concept, it is found that the proportion of Protestants that were calm for 2 out of the 3 days is of 0.0606.
What is a proportion?A proportion is a fraction of total amount, and is given by the number of desired outcomes divided by the number of total outcomes.In this problem:
There is a total of 50 + 27 + 6 + 16 = 99 Protestants.Of those, 6 were calm for 2 out of the 3 days.Hence:
[tex]p = \frac{6}{99} = 0.0606[/tex]
The proportion of Protestants that were calm for 2 out of the 3 days is of 0.0606.
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Approximately 6.06% of Protestants were calm for 2 out of the 3 days, calculated by dividing the number of Protestants calm for 2 days (6) by the total surveyed (99).
Explanation:To find the proportion of Protestants who were calm for 2 out of the 3 days, we need to look at the numbers given for Protestant individuals and calculate the ratio of those who were calm for exactly two days to the total number of Protestants surveyed. According to the data provided:
Number of Protestants calm for 0 days: 50Number of Protestants calm for 1 day: 27Number of Protestants calm for 2 days: 6Number of Protestants calm for 3 days: 16The total number of Protestants surveyed is the sum of those calm for 0, 1, 2, and 3 days, which is 50 + 27 + 6 + 16 = 99.
To find the proportion of Protestants calm for 2 days, we divide the number calm for 2 days by the total number of Protestants surveyed:
Proportion = Number calm for 2 days / Total number of Protestants
Proportion = 6 / 99
Proportion ≈ 0.0606
This means that approximately 6.06% of Protestants were calm for 2 out of the 3 days.
Suppose that diameters of a new species of apple have a bell-shaped distribution with a mean of 7.46cm and a standard deviation of 0.39cm. Using the empirical rule, what percentage of the apples have diameters that are greater than 7.07cm? Please do not round your answer.
Using the empirical rule, approximately 84% of apple diameters are greater than 7.07 cm, since this value is within one standard deviation below the mean diameter of 7.46 cm.
Explanation:The empirical rule (also known as the 68-95-99.7 rule) is a statistical rule which states that for a normally distributed set of data, approximately 68% of data falls within one standard deviation of the mean, 95% within two standard deviations, and 99.7% within three standard deviations.
In this particular problem, we know the mean (7.46cm) and standard deviation (0.39cm) of the apple diameters. We are asked to find the percentage of apple diameters that are greater than 7.07cm. Since 7.07cm is less than one standard deviation below the mean (7.46 - 0.39 = 7.07), from the normal distribution we can say that approximately 84% (50% of the data is above the mean and 34% is between the mean and one standard deviation below the mean) of the apple diameters are larger than 7.07cm according to the Empirical rule.
Thus, using Empirical rule and given statistical mean and standard deviation, we can estimate that about 84% of the apples from this species have a diameter greater than 7.07 cm.
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Using the Empirical Rule, approximately 84% of the apples have diameters greater than 7.07cm since 7.07cm is exactly one standard deviation below the mean of 7.46cm in a normal distribution.
To answer the student's question, we need to apply the Empirical Rule which is used in statistics to describe the distribution of data in a bell-shaped curve, also known as a normal distribution. This rule states that for a data set with a normal distribution, approximately 68% of the data falls within one standard deviation from the mean, 95% falls within two standard deviations, and over 99% falls within three standard deviations.
Given that the mean diameter of the apples is 7.46cm and the standard deviation is 0.39cm, first we need to determine how many standard deviations 7.07cm is from the mean. This is calculated as:
(7.07cm - 7.46cm) / 0.39cm = -1 standard deviation.
According to the Empirical Rule, 68% of apples are within one standard deviation above and below the mean. Since 7.07cm falls exactly at one standard deviation below the mean, we have:
50% of the apples have diameters greater than the mean.Another 34% (half of 68%) have diameters that fall between the mean and one standard deviation below the mean.Therefore, to find the percentage of apples that have diameters greater than 7.07cm, we would add these two percentages together:
50% + 34% = 84%
Thus, we conclude that approximately 84% of the apples have diameters that are greater than 7.07cm.
The revenue (in millions of dollars) from the sale of x units at a home supply outlet is given by R(x) = .21x. The profit (in millions of dollars) from the sale of x units is given by P(x) = .084x - 1.5. a. Find the cost equation. b. What is the cost of producing 7 units? c. What is the break-even point?
Answer: a) C = 0.126x + 1.5
b) C = 2.382 c) approximately 179 units
Step-by-step explanation: profit P(x), total revenue R(x) and total cost (C) are related by the formulae below.
Profit = total revenue - total cost
P(x) = R(x) - C
From the question, P(x) = 0.084x - 1.5, R(x) = 0.21x.
Question a)
C = R(x) - P(x)
C = 0.21x - {0.084x - 1.5}
C = 0.21x - 0.084x + 1.5
C = 0.126x + 1.5
Question b)
If C = 0.126x + 1.5, then C at x = 7 units is gotten below as
C = 0.126(7) + 1.5
C = 0.882 + 1.5
C = 2.382.
Question c)
The break even point is the point where total revenue R(x) equals total cost C.
R(x) = 0.21x and C = 0.126x + 1.5
0.21x = 0.126x + 1.5
0.21x - 0.126x = 1.5
0.084x = 15
x = 15/ 0.084
x = 178.57 which is approximately 179 units (since quantity of units can't be decimal)
To find the cost equation, subtract the profit equation from the revenue equation. The cost of producing 7 units is approximately 2.382 million dollars. The break-even point occurs at approximately 17.857 units.
Explanation:a. To find the cost equation, we need to subtract the profit equation from the revenue equation. Since the revenue equation is R(x) = .21x and the profit equation is P(x) = .084x - 1.5, the cost equation is C(x) = R(x) - P(x). Substituting the given equations, we have C(x) = .21x - (.084x - 1.5).
Simplifying the equation, we get C(x) = .21x - .084x + 1.5.
Combining like terms, the cost equation is C(x) = .126x + 1.5.
b.
To find the cost of producing 7 units, we can substitute x = 7 into the cost equation. C(7) = .126(7) + 1.5 = .882 + 1.5 = 2.382 million dollars.
c.
The break-even point occurs when the revenue equals the cost. To find the break-even point, we set R(x) = C(x). Substituting the given equations, we have .21x = .126x + 1.5.
Solving for x, we get .084x = 1.5.
Dividing both sides by .084, we find x = 17.857 units.
Therefore, the break-even point is approximately 17.857 units.
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Consider this scatter plot.
(a) How would you characterize the relationship between the hours spent on homework and the test
scores? Explain.
(b) Paul uses the function y = 8x + 42 to model the situation. What score does the model predict for 3 h
of homework?
Answer:
a) For this case we can see on the y axis the test scores and on the x axis the hours of homework. And as we can see we have a proportional relationship, because when the hours of homework increase the test scores increases too. If we fit a lineal model between y and x we will see an slope positive and a correlation coefficient positive based on the data observed.
b) [tex] y = 8x +42[/tex]
And we want to predict the test score for x = 3hr of homework we just need to replace the value of x=3 in the linear model and we got:
[tex] y = 8*3 +42= 24+42=66[/tex]
And that would be our predicted value for 3 h of homework
Step-by-step explanation:
For this case we consider the scatter plot attached to solve the problem.
Part a
For this case we can see on the y axis the test scores and on the x axis the hours of homework. And as we can see we have a proportional relationship, because when the hours of homework increase the test scores increases too. If we fit a lineal model between y and x we will see an slope positive and a correlation coefficient positive based on the data observed.
Part b
Assuming the following linear model for the situation:
[tex] y = 8x +42[/tex]
And we want to predict the test score for x = 3hr of homework we just need to replace the value of x=3 in the linear model and we got:
[tex] y = 8*3 +42= 24+42=66[/tex]
And that would be our predicted value for 3 h of homework
A perpetuity pays $50 per year and interest rates are 9 percent. How much would its value change if interest rates decreased to 6 percent?
Answer:
We conclude that its value change for 277.8$.
Step-by-step explanation:
We know that a perpetuity pays $50 per year and interest rates are 9 percent. We calculate how much would its value change if interest rates decreased to 6 percent. We know that
9%=0.09
6%=0.06
We get
\frac{50}{0.09}=555.5
\frac{50}{0.06}=833.3
Therefore, we get 833.3-555.5=277.8
We conclude that its value change for 277.8$.
The value change for $277.8.
The calculation is as follows:
[tex]\frac{50}{0.09}=555.5\\\\\frac{50}{0.06}=833.3[/tex]
So,
= $833.3 - $555.5
= $277.8
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(10 pts) A device has a sensor connected to an alarm system. The sensor triggers 95% of the time if dangerous conditions exist and 0.5% of the time if conditions are normal. Dangerous conditions exist 0.5% of the time in general. (a) What is the probability of a false alarm
Answer:
The probability of a false alarm is 0.5116 .
Step-by-step explanation:
Let us indicate three events :
Event A = Alarm system triggers
Event B = Dangerous conditions exist
Event B' = Normal conditions exist
Now we are given with P(A/B) which means probability of alarm getting triggered given the dangerous conditions exist , P(A/B') which means probability of alarm getting triggered given the normal conditions exist and P(B) which means the probability of dangerous conditions existing i.e.
P(A/B) = 0.95 P(A/B') = 0.005 P(B) = 0.005 P(B') = 1 - P(B) = 0.995
(a) The probability of a false alarm means given the alarm gets triggered probability that there was normal conditions i.e. P(B'/A)
P(B'/A) = [tex]\frac{P(A\bigcap B')}{P(A)}[/tex]
Now P(A) = P(B) * P(A/B) + P(B') * P(A/B') {This representing Probability of alarm getting triggered in both the conditions]
P(A) = 0.005 * 0.95 + 0.995 * 0.005 = 9.725 x [tex]10^{-3}[/tex]
Since P(A/B') = 0.005
[tex]\frac{P(A\bigcap B')}{P(B')}[/tex] = 0.005 So, [tex]P(A\bigcap B')[/tex] = 0.005 * 0.995 = 4.975 x [tex]10^{-3}[/tex]
Therefore, P(B'/A) = [tex]\frac{P(A\bigcap B')}{P(A)}[/tex] = [tex]\frac{4.975*10^{-3} }{9.725*10^{-3} }[/tex] = 0.5116 .
Disks of polycarbonate plastic from a supplier are analyzed for scratch and shock resistance. The results from 100 disks are summarized as follows: Shock resistance High LowScratch High 70 9Resistance Low 16 5Let A denote the event that a disk has high shock resistance, and let B denote the event that a disk has high scratch resistance. Are events A and B independent?
Answer:
Part a
The probability that the event disk has high shock resistance is 0.86
Part b
The probability that a disk has high scratch resistance given that the disk has high shock resistance is 0.8140
Step-by-step explanation:
(a).
From the given information,
Let A denote the event that a disk has high shock resistance,
and let B denote the event that a disk has high scratch resistance.
SHOCK HIGH(A) SHOCK LOW(A) TOTAL
SCRATCH
HIGH(B) 70 9 79
RESISTANCE 16 5 21
LOW(B)
TOTAL 86 14 100
Compute P(A).
Therefore, the probability value of the event A is 0.86.
Part a
The probability that the event disk has high shock resistance is 0.86
Explanation | Hint for next step
Based on the given information, the probability that the event disk has high shock resistance is 0.86. That means it is approximately equal to 86%.
Step 2 of 2
(b)
From the given information,
Let A denote the event that a disk has high shock resistance,
and let B denote the event that a disk has high scratch resistance.
SHOCK HIGH(A) SHOCK LOW(A) TOTAL
SCRATCH
HIGH(B) 70 9 79
SCRATCH
LOW(B) 16 5 21
TOTAL 86 15 100
ComputeP(B/A) = P(A∩B) /P(A) ;P(A) > 0
P(A∩B) =70/100
=0.70
From the part [a], the probability value of the event A is P(A) =0.86 .
Therefore,
P( {B/A} = P(A∩B) /P(A)
= 0.70 /0.86
=0.8140
Part b
The probability that a disk has high scratch resistance given that the disk has high shock resistance is 0.8140
Explanation | Common mistakes
The disk with high scratch resistance is found to be 81.40% with the condition that the disk with high shock resistance is maintained.
Final answer:
Events A (high shock resistance) and B (high scratch resistance) in the context of polycarbonate plastic disks are not independent, as the calculated probabilities P(A ∩ B) and P(A)P(B) do not match.
Explanation:
Whether two events are independent can be determined by checking if the probability of one event occurring does not affect the probability of the other event. Mathematically, two events A and B are independent if and only if-
P(A ∩ B) = P(A)P(B)
Let's calculate this using the data provided from the analysis of 100 disks of polycarbonate plastic.
Total number of disks (n) = 100
Number of disks with high shock resistance (A) = 70 + 9 = 79
Number of disks with high scratch resistance (B) = 70 + 16 = 86
Number of disks with both high shock and scratch resistance (A ∩ B) = 70
The probability of A (P(A)) is thus 79/100, the probability of B (P(B)) is 86/100, and the probability of both A and B occurring (P(A ∩ B)) is 70/100. Now we multiply P(A) and P(B):
P(A)P(B) = (79/100) × (86/100) = 0.6794
However, P(A ∩ B) as observed is-
70/100 = 0.7
Since P(A ∩ B) does not equal P(A)P(B), the events A and B are not independent.
1. Rewrite each condition below in valid Java syntax (give a boolean expression): a. x > y > z b. x and y are both less than 0 c. neither x nor y is less than 0 d. x is equal to y but not equal to z
Answer:
Here are Boolean expressions in Java syntax.
Step-by-step explanation:
For each part:
a) (x > y && y > z)
b) x == y && x < 0, or x < 0 && y < 0, or x == y && y < 0 (which is essentially the first example)
c) (x == y && x >= 0), or (x >= 0 && y >= 0), or (x == y && y >= 0) (for the first and third, once the first condition establishes that x is equal to y, if either x or y is greater than or equal to 0, then they are both not less than 0)
d) (x == y && x != z), or (x == y && y != z)
The binomial formula has two parts. The first part of the binomial formula calculates the number of combinations of X successes. The second part of the binomial formula calculates the probability associated with the combination of success and failures. If N=6 and X=4, what is the number of combinations of X successes?156486!
Answer:
15 is the number of combination of 4 successes.
Step-by-step explanation:
We are given the following information:
We are given a binomial distribution, then probability of x succes is given by
[tex]P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}[/tex]
where n is the total number of observations, x is the number of success, p is the probability of success.
Now, we are given n = 6 and x = 4
We have to evaluate the number of combination of success and failures.
It is given by:
[tex]\binom{n}{x} = \dfrac{n!}{x!(n-x)!}\\\\\binom{6}{4} = \dfrac{6!}{4!(6-4)!} = \dfrac{6!}{4!2!} = 15[/tex]
Thus, 15 is the number of combination of 4 successes.
The number of combinations of 4 successes out of 6 attempts, calculated using the binomial probability combination formula, is 15.
Explanation:In binomial probability, the number of ways to get a specific number of successes is often calculated using the combination formulas. In this case, the number of combinations of X successes would be determined by the combination formula C(n, x), which stands for the number of combinations of n items taken x at a time. In your case, where n=6 and x=4, the required number of combinations (or ways to achieve 4 successes out of 6 trials) is C(6,4).
The formula for a combination is generally given by: n! / [x!(n - x)!]. Plugging the given numbers into the formula, we get: C(6,4) = 6! / [4!(6 - 4)!]. This simplifies to: 720 / (24*2), which equals to 15. So, there are 15 combinations of 4 successes out of 6 attempts.
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