Answer: 0.0136
Step-by-step explanation:
Let events are:
A = alarm A will fail
B= alarm B will fail
We have given ,
P(A) = 0.004 , P(B)=0.01
Since both events are independent , so
P(A and B) = P(A) x P(B)
= 0.04 x 0.01 =0.0004
i.e. P(A and B) =0.0004
Now , P(A or B) = P(A)+P(B)-P(A and B)
= 0.004+ 0.01-0.0004=0.0136
Hence, the probability that at least one alarm fails in the event of a fire is 0.0136 .
The probability that at least one alarm fails in the event of a fire is approximately 0.014 or 1.4%.
The probability that at least one alarm fails in the event of a fire is given by:
[tex]\[ P(\text{at least one fails}) = 1 - P(\text{both work}) \][/tex]
Since the failures are independent, the probability that both alarms work is the product of their individual probabilities of working:
[tex]\[ P(\text{both work}) = P(\text{A works}) \times P(\text{B works}) \][/tex]
The probability that alarm A works is the complement of the probability that it fails, which is [tex]\( 1 - 0.004 \)[/tex]. Similarly, the probability that alarm B works is 1 - 0.01 .
Thus, we have:
[tex]\[ P(\text{A works}) = 1 - 0.004 = 0.996 \] \[ P(\text{B works}) = 1 - 0.01 = 0.99 \][/tex]
Now we can calculate [tex]\( P(\text{both work}) \)[/tex]:
[tex]\[ P(\text{both work}) = 0.996 \times 0.99 \] \[ P(\text{both work}) = 0.98604 \][/tex]
Finally, we find the probability that at least one alarm fails:
[tex]\[ P(\text{at least one fails}) = 1 - 0.98604 \] \[ P(\text{at least one fails}) = 0.01396 \][/tex]
A farmer caught some wild pigs to use as founders for a domestic herd. The farmer wants all of her pigs to have one distinctive color. Most of the wild pigs are brown, but she catches a few with red hair and a few with black hair. The farmer plans to breed either the black-haired or the red-haired pigs to establish a pure-breeding herd (eliminating all other colors). Black hair (B) is dominant over brown (W). Black (B) and brown (W) are dominant over red (R). If the farmer wants her whole herd to reach its new color in as few generations as possible, should she breed for red-haired pigs, or should she breed for black-haired pigs? Why?
Answer: Breed for black-haired pigs
Step-by-step explanation:
Now according to the statement, the farmer catches wild pigs and mostly are brown.
Since black hair is dominant over brown and also black and brown are dominant over red, this implies that red is the least dominant color meaning if breeding occurs, there would be less chance of getting red colored pigs.
She wants to breed either black or red haired pigs and since black is more dominant so she should breed the black ones with the brown ones or with themselves so she would eventually get pure-breeding herd i.e. of black color.
Suppose that the distance, in miles, that people are willing to commute to work is an exponential random variable with a decay parameter \frac{1}{20}. Let X = the distance people are willing to commute in miles. What is m, μ, and σ? What is the probability that a person is willing to commute more than 25 miles?
Answer:
m = [tex]\frac{1}{20}[/tex] μ = 20σ = 20The probability that a person is willing to commute more than 25 miles is 0.2865.
Step-by-step explanation:
Exponential probability distribution is used to define the probability distribution of the amount of time until some specific event takes place.
A random variable X follows an exponential distribution with parameter m.
The decay parameter is, m.
The probability distribution function of an Exponential distribution is:
[tex]f(x)=me^{-mx}\ ;\ m>0, x>0[/tex]
Given: The decay parameter is, [tex]\frac{1}{20}[/tex]
X is defined as the distance people are willing to commute in miles.
The decay parameter is m = [tex]\frac{1}{20}[/tex]. The mean of the distribution is: [tex]\mu=\frac{1}{m}=\frac{1}{\frac{1}{20}}=20[/tex]. The standard deviation is: [tex]\sigma=\sqrt{variance}= \sqrt{\frac{1}{(m)^{2}} } =\frac{1}{m} =\frac{1}{\frac{1}{20}} =20[/tex]Compute the probability that a person is willing to commute more than 25 miles as follows:
[tex]P(X>25)=\int\limits^{\infty}_{25} {\frac{1}{20} e^{-\frac{1}{20}x}} \, dx \\=\frac{1}{20}|20e^{-\frac{1}{20}x}|^{\infty}_{25}\\=|e^{-\frac{1}{20}x}|^{\infty}_{25}\\=e^{-\frac{1}{20}\times25}\\=0.2865[/tex]
Thus, the probability that a person is willing to commute more than 25 miles is 0.2865.
how many pounds of bananas cost $3.75
Answer:
you go up by .75 each time per pound so it would be 5 pounds.
hope this helps!!
Answer:
5
Step-by-step explanation:
Or (A)
a fair die is tossed. A is the event that the outcome is odd. B is the event that the outcome is even. C is the event that the outcome is ess than 4.
Answer:
The question is incomplete. Below is the complete question
"A fair die is tossed. A is the event that the outcome is odd. B is the event that the outcome is even. C is the event that the outcome is less than 4. Determine P(A), P(B), P(C), P(A∩B), P(B∩C), P(A|B), P(A|C), P(B|C).
answers:
a.1/2
b. 1/2
c. 1/2
d.0
e. 1/6
f. 0
g. 2/3
h. 1/3
Step-by-step explanation:
lets write out the probability of each event
a.A=outcome is odd
A={1,3,5}
P(A)=1/2
b.B=outcome is even
B={2,4,6}
P(B)=1/2
c. C=outcome is less than 4
C={1,2,3}
P(C)=1/2
d.P(AnB)={its odd and even i.e what is common between set A and B}
Therefore, P(AnB) is a null set i.e P(AnB)={ }=0
e. BnC= {2} i.e elements common between set B and C
Hence P(BnC)=1/6
f. P{A|B}= P(AnB)/P(B) since P(AnB)=0, P(B)=1/2. then, P(A|B)=0
g. P(A|C)= P(AnC)/P(C) since P(AnC)=1/3, P(C)=1/2. therefore, P(A|C)=2/3
h. P(B|C)= P(BnC)/P(C), since P(BnC)=1/6, P(C)=1/2. therefore, P(B|C)=1/3
This person has a 75% chance of a full recovery. Is this classical probability, empirical or subjective
Answer:
Subjective probability
Step-by-step explanation:
At first we should know the following:
Classical probability ⇒ when there are n equally likely outcomes.
Subjective probability ⇒ is based on whatever information is available.
Empirical probability ⇒ when the number of times the event happens is divided by the number of observations.
So, according to the previous definitions:
This person has a 75% chance of a full recovery
There is no equally likely outcomes, and the percentage of full recovery is based on the information available about the person and also it is based on educated guess.
So, this is Subjective probability
A faculty leader was meeting two students in Paris, one arriving by train from Amsterdam and the other arriving from Brussels at approximately the same time. Let A and B be the events that the trains are on time, respectively. If P(A) = 0.93, P(B) = 0.89 and P(A \ B) = 0.87, then find the probability that at least one train is on time.
Answer: P(AUB) = 0.93 + 0.89 - 0.87 = 0.95
Therefore, the probability that at least one train is on time is 0.95.
Step-by-step explanation:
The probability that at least one train is on time is the probability that either train A, B or both are on time.
P(A) = P(A only) + P(A∩B)
P(B) = P(B only) + P(A∩B)
P(AUB) = P(A only) + P(B only) + P(A∩B)
P(AUB) = P(A) + P(B) - P(A∩B) ......1
P(A) = 0.93
P(B) = 0.89
P(A∩B) = 0.87
Then we can substitute the given values into equation 1;
P(AUB) = 0.93 + 0.89 - 0.87 = 0.95
Therefore, the probability that at least one train is on time is 0.95.
A pack of Original Starbursts contains four colors/flavors: red (cherry), orange (orange), yellow (lemon), and pink (strawberry). A recent survey asked people what their favorite Starburst color (or flavor) is. They asked 756 people and recorded their favorite color (or flavor) and their age group. The table below provides the results:
What is your favorite Original Starburst color (flavor)? Oran Red (C 81 69 75 225 Group e) Yellow (Lemon Pink (Stra 124 121 114 359 Total 254 252 250 756 ildren (6-12 years old) Teenagers (13- 19 years old) 25 32 28 85 24 30 ng Adults (20 - 26 years old) otal 87
What is the probability that a randomly selected survey respondent is a Teenager OR selects Pink (Strawberry) as their favorite Starburts color (flavor)?
0.1601
0.1735
0.3333
0.4749
0.6481
0.6548
0.6667
Based on all your calculations, the events "a randomly selected person is a Teenager" and 'a randomly selected person selects Pink (Strawberry) as their favorite Starburst color (flavor)" would be considered (check all that applies):
dependent events
independent events
mutually exclusive events
complementary events
You could determine if the events "a randomly selected person is a Teenager" and "a randomly selected person selects Pink (Strawberry) as their favorite Starburst color (flavor)" were independent by comparing the P(Pink | Teenager) with which of the following probabilities?
0.1601
0.4560
0.4749
0.4882
0.5000
0.6481
0.6667
Answer:
P ( P U T ) = 0.6481
Dependent events
P ( P / T ) = (359/756) = 0.4749
Step-by-step explanation:
Given:
- Pink : P
- Teenagers: T
Find:
- P( T u P )
- Are the events T and R dependent, independent, mutually exclusive, or complementary
- If the events are independent, then we can compare P ( P / T ).
Solution:
a)
For first part we will determine the total outcome for T and P:
Total outcomes P or T = 359 + 252 - 121 = 490
(P U T) = (P) + (T) - (P n T)
The total number of possible outcomes are = 756
Hence, P ( P U T ) = 490 / 756 = 0.6481
b)
We are to investigate how the two events are related to one another:
- Check for dependent events:
P ( T n P ) = P( T ) * P ( P / T )
(121 / 756 ) = (252/756) * ( 121 / 252 )
(121 / 756) = (121/756) ....... Hence, events are dependent
- Check for independent events:
P ( T n P ) = P( T ) * P ( P )
(121 / 756 ) = (252/756) * ( 359 / 756 )
(121 / 756) =/ (359/756) ....... Hence, events are not independent
- Check for mutually exclusive events:
P ( T U P ) = P( T ) + P ( P )
(490 / 756 ) = (252/756) + ( 359 / 756 )
(121 / 756) =/ (611/756) ... Hence, events are not mutually exclusive
Hence, the two events are dependent on each other.
c)
If the events are said to be independent then the event:
P ( P / T ) = P (P)
P ( P / T ) = (359/756) = 0.4749
The probability that a randomly selected survey respondent is a Teenager OR selects Pink (Strawberry) as their favorite Starburst color (flavor) can be calculated by adding the individual probabilities of these two events occurring.
The probability of a respondent being a Teenager is 85/756, and the probability of selecting Pink (Strawberry) as their favorite flavor is 359/756. So, the probability of either event happening is (85/756) + (359/756) = 0.4749.
Based on these calculations, the events "a randomly selected person is a Teenager" and "a randomly selected person selects Pink (Strawberry) as their favorite Starburst color (flavor)" are NOT mutually exclusive events because they can both happen.
They are also NOT complementary events because they don't cover all possible outcomes.
Whether they are independent or dependent events can be determined by comparing P(Pink | Teenager) with the probability of Pink (Strawberry) regardless of age group, P(Pink).
If P(Pink | Teenager) is equal to P(Pink), they are independent events; otherwise, they are dependent.
In this case, P(Pink | Teenager) is 114/85, and P(Pink) is 359/756. Comparing these probabilities, we find that P(Pink | Teenager) is not equal to P(Pink), so the events are dependent.
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1. A shipping company operates 10 container ships that can each carry 5000 containers on each journey between ports. They want to be able to load and unload 20,000 containers each week. Assume their ships always travel fully loaded. What is the longest average travel time between ports that allows them to meet their goal of 20,000 containers per week
Answer:
2.5 weeks
Step-by-step explanation:
Data provided in the question:
Number of container ships operated by the company = 10
Number of containers carried by each container = 5000
Number of container to be unloaded = 20,000
A shipping company operates 10 container ships
Now,
Capacity of 10 container ships = 10 × 5000 containers = 50,000 containers
Time required to load and upload 50,000 containers
= 50,000 containers ÷ 20,000 containers per week
= 2.5 weeks
Determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist. (Enter your answer using interval notation.) (t − 5)y' + (ln t)y = 6t, y(1) = 6
Answer:
0 < t < 5 is the required interval for the differential equation (t - 5)y' + (ln t)y = 6t to have a solution.
Step-by-step explanation:
Given the differential equation
(t - 5)y' + (ln t)y = 6t
and the condition y(1) = 6
We can rewrite the differential equation by dividing it by (t - 5) as
y' + [(ln t)/(t - 5)]y = 6t/(t - 5)
(ln t)/(t - 5) is continuous on the interval (0, 5) and (5, +infinity).
6t/(t - 5) is continuous on (-infinity, 5) and (5, +infinity)
We see that for these expressions, we have continuity at the intervals (0, 5) and (5, +infinity).
But the initial condition is y = 6, when t = 1.
The solution to differential equation is certain to exist at (0, 5)
Which implies that
0 < t < 5
is the required interval.
How much work is required to lift a 1400-kg satellite to an altitude of 2⋅106 m above the surface of the Earth? The gravitational force is F=GMm/r2, where M is the mass of the Earth, m is the mass of the satellite, and r is the distance between the satellite and the Earth's center. The radius of the Earth is 6.4⋅106 m, its mass is 6⋅1024 kg, and in these units the gravitational constant, G, is 6.67⋅10−11.
To lift a 1400-kg satellite to an altitude of 2×10⁶ meters above the Earth's surface, the work required is approximately 2.079×10¹¹ Joules. This is calculated based on changes in gravitational potential energy using given values for mass, radius, and the gravitational constant.
To calculate the work required to lift a 1400-kg satellite to an altitude of 2×10⁶ meters (2000 km) above the Earth's surface, we need to consider the gravitational potential energy change.
Given:
Mass of the satellite (m): 1400 kgAltitude above Earth's surface (h): 2×10⁶ mRadius of the Earth (Rₑ): 6.4×10⁶ mMass of the Earth (M): 6×10²⁴ kgGravitational constant (G): 6.67×10⁻¹¹ N·m²/kg²The total distance from the center of the Earth to the satellite is:
r = Rₑ + h = 6.4×10⁶ m + 2×10⁶ m = 8.4×10⁶ m.The work required is equal to the change in gravitational potential energy:
The gravitational potential energy at a distance r from the Earth’s center is given by:
U = -GMm/rThe work done (W) to move the satellite from the Earth's surface to this altitude is the difference in potential energy:
W = GMm (1/Rₑ - 1/r)Substitute the given values:
W = (6.67×10⁻¹¹ N·m²/kg²)(6×10²⁴ kg)(1400 kg) [(1/6.4×10⁶ m) - (1/8.4×10⁶ m)]Calculate the values inside the brackets first:
(1/6.4×10⁶ - 1/8.4×10⁶) ≈ 1.5625×10⁻⁷ - 1.1905×10⁻⁷ ≈ 0.372×10⁻⁷Now, multiply:
W ≈ 6.67×10⁻¹¹ × 6×10²⁴ × 1400 × 0.372×10⁻⁷W ≈ 2.079×10¹¹ JoulesThe work required to lift the satellite to the desired altitude is approximately 2.079×10¹¹ Joules.
Each of the following passages contains a single argument. Using the letters "P" and "C," identify the premises and conclusion of each argument, writing premises first and conclusion last. List the premises in the order in which they make the most sense (usually the order in which they occur), and write both premises and conclusion in the form of separate declarative sentences. Indicator words may be eliminated once premises and conclusion have been appropriately labeled. An ant releases a chemical when it dies, and its fellows then carry it away to the compost heap. Apparently the communication is highly effective; a healthy ant painted with the death chemical will be dragged to the funeral heap again and again.
Answer:
P1: An ant releases a chemical when it dies, and its fellows then carry it away to the compost heap.
P2: A healthy ant painted with the death chemical will be dragged to the funeral heap again and again.
C: Apparently the communication is highly effective
Step-by-step explanation:
Premises are statements that are always assumed to be truth or statements that have been given and made to be true.
Premises will be labelled as P1, P2, P3..........
Conclusions are derived statements from premises. Its truthfulness is derived from the truthfulness of premises. Meaning that, when the premises is true, the conclusion is also true.
Conclusions will be labelled as C1, C2, C3........
In the question above, it's noted that the conclusion starts at "Apparently the" while the premises are statements before "Apparently the"
The argument's premises are as follows:
1. An ant releases a chemical when it dies.
2. Its fellow ants then carry the dead ant to the compost heap.
3. A healthy ant painted with the death chemical will be dragged to the funeral heap again and again.
The conclusion is that the communication among ants when one of them dies is highly effective.
In the given passage, there is an implicit argument about the effectiveness of communication among ants when one of them dies. Let's identify the premises and the conclusion:
Premises:
1. An ant releases a chemical when it dies.
2. Its fellow ants then carry the dead ant to the compost heap.
3. A healthy ant painted with the death chemical will be dragged to the funeral heap again and again.
Conclusion:
4. Apparently, the communication among ants when one of them dies is highly effective.
These premises and conclusion can be organized as follows:
Premise 1: An ant releases a chemical when it dies.
Premise 2: Its fellow ants then carry the dead ant to the compost heap.
Premise 3: A healthy ant painted with the death chemical will be dragged to the funeral heap again and again.
Conclusion: Apparently, the communication among ants when one of them dies is highly effective.
In this argument, premises 1, 2, and 3 provide evidence or observations related to ant behavior when a member of their colony dies. The conclusion, stated in premise 4, is drawn from these observations and suggests that the communication system among ants regarding the death of a fellow ant is effective.
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SOLUTION: if you know the names of the remaining 5 students in the spelling bee, what is the probability of randomly selecting an order and getting the order that is used in the spelling bee
There are
[tex]5! = 5\cdot 4 \cdot 3 \cdot 2 = 120[/tex]
possible arrangements of 5 students. So, if you pick a particular one, you'll have a probability of 1/120 to guess the correct one.
The probability of getting the order used in the spelling bee is 1/120 or approximately 0.0083.
Explanation:The probability of randomly selecting an order and getting the order used in the spelling bee is 1 divided by the number of possible orders. To find the number of possible orders, we use the concept of permutations. If there are 5 students participating in the spelling bee, there are 5 factorial (5!) possible orders. So the probability is 1/5! which is equal to 1/120 or approximately 0.0083.
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Genome expression profiles and drug design: reseracher are trying to design a druq that alters the expression level of the three genes in a specific way. The desired change in expression, written as a vector in V3 is [3,9,-5], where all measurment are dimentionless. suppose two potential drugs are being studied, and the expression profiles that they induce are (A) [2,4,1] and (B) [-2,3,-5]. (a) What is the magnitude of the desired change in expression? (b) USing scalar projections, determine the fraction of this desired change that is induced by each drug. (c) Which drug is closest to having the desired effect?
Answer:
a) Magnitude = 10.72
b) = 0.32 and 0.400
c) Drug B is closest to having the desired effect.
Step-by-step explanation:
The knowledge of vector projection and dot product is applied as shown in the step by step explanation from the attached file.
Solve, graph, and give interval notation for the compound inequality:
−4x + 1 > 13 AND 4(x + 2) ≤ 4
Answer:
The answer to your question is below
Step-by-step explanation:
Inequality 1
-4x + 1 > 13
-4x > 13 - 1
-4x > 12
x < 12/-4
x < -3
Inequality 2
4(x + 2) ≤ 4
4x + 8 ≤ 4
4x ≤ 4 - 8
4x ≤ - 4
x ≤ -4/4
x ≤ -1
Interval notation (-∞, -3)
See the graph below
The solution to these inequalities in where both intervals crosses
According to a study done by Nick Wilson of Otago University Wellington, the probability a randomly selected individual will not cover his or her mouth when sneezing is 0.267. Suppose you sit on a bench in a mall and observe people's bad habits as they sneeze.a. What is the probability that among 10 randomly observed individuals exactly 4 do not cover their mouth when sneezing?b. What is the probability that among 10 randomly observed individuals fewer than 3 do not cover their mouth when sneezing?c. Would you be surprised if, after observing 18 individuals, fewer than half covered their mouth when sneezing? why?
Answer and Step-by-step explanation:
From the question statement we get know that it is Binomial distribution because there are only two possible outcomes so we need to use Binomial Probability Distribution for this question.
Formula for the Binomial Probability Distribution:
P(X)= p^x q^(n-x)
Where,
C_x^n=n!/(n-x)!x! (i.e. combination)x= total number of successes p=probability of success (p=1-q) q=probability of failure (q=1-p) n=number of trials P(X)= probability of total number of successesAnswer and explanation for each part of the question are as follow:
a.What is the probability that among 10 randomly observed individuals exactly 4 do not cover their mouth when sneezing?
Solution:
Given that
n=10
p=0.267 (because p is the probability of success which is “number of individuals not covering their mouths when sneezing” in the question)
q=1-0.267=0.733
x=4 (number of successes i.e. individuals not covering their mouths)
C_x^n=n!/(n-x)!x!=10!/(10-4)!4!=210
P(X)=C_x^n p^x q^(n-x)=210×〖(0.267)〗^4×〖0.733〗^(10-4)
P(X)=210×0.00508×0.155
P(X)=0.165465
b. What is the probability that among 10 randomly observed individuals fewer than 3 do not cover their mouth when sneezing?
Solution:
Given that
n=10
p=0.267 (because p is the probability of success which is “number of individuals not covering their mouths when sneezing” in the question)
q=1-0.267=0.733
x=3 (number of successes i.e. individuals not covering their mouths)
C_x^n=n!/(n-x)!x!=10!/(10-3)!3!=120
P(X)=C_x^n p^x q^(n-x)=120×(0.267)^3×〖0.733〗^(10-3)
P(X)=120×0.01903×0.1136
P(X)=0.25962
c. Would you be surprised if, after observing 18 individuals, fewer than half covered their mouth when sneezing? why?
Solution:
Given that
n=18
p=0.267 (because p is the probability of success which is “number of individuals not covering their mouths when sneezing” in the question)
q=1-0.267=0.733
x=9 (x is the number of successes “number of individuals not covering their mouths when sneezing”, if less than half cover their mouth then more than half will not cover), so let x=9
C_x^n=n!/(n-x)!x!=18!/(18-9)!9!=48620
P(X)=48620×(0.267)^9×〖0.733〗^(18-9)
P(X)=48620×0.00000689×0.0610
P(X)=0.020
Yes, I am surprised that probability of less than 9 individuals covering their mouth when sneezing is 0.020. Which is extremely is small.
How many equations can you write that will equal 100
Step-by-step explanation:
Infinitely many equations can be written that will be equal to 100.
x + y = 100
2x - y = 100
and many more..
If a person rolls a six dash sided dierolls a six-sided die and then flips a coinflips a coin, describe the sample space of possible outcomes using 1 comma 2 comma 3 comma 4 comma 5 comma 61, 2, 3, 4, 5, 6 for the diedie outcomes and Upper H comma Upper TH, T for the coincoin outcomes. (Make sure your answers reflect the order stated.) The sample space is Sequals=_________{nothing}.
Answer:
S={1H,2H,3H,4H,5H,6H,1T,2T,3T,4T,5T,6T}
Step-by-step explanation:
The sample space for a six sided die is {1,2,3,4,5,6} as there are 6 possible outcomes and for flipping a coin is {H,T} as there are two possible outcomes.
If a person rolls a six sided die and then flips a coin then the sample space for this event will be written as
S={1H,2H,3H,4H,5H,6H,1T,2T,3T,4T,5T,6T}.
The number of elements in the sample space are
n(S)=12.
Answer:
Step-by-step explanation:If a person draws a playing card and checks its color and then spins a six dash space spinner, describe the sample space of possible outcomes using Upper B comma Upper R for the card outcomes and 1 comma 2 comma 3 comma 4 comma 5 comma 6 for the spinner outcomes. (Make sure your answers reflect the order stated.)
The sample space is Sequals{
012
12}.
(Use a comma to separate answers as needed.)
Because of safety considerations, in May 2003 the Federal Aviation Administration (FAA) changed its guidelines for how small commuter airlines must estimate passenger weights. Under the old rule, airlines used 180 lb as a typical passenger weight (including carry-on luggage) in warm months and 185 lb as a typical weight in cold months. The Alaska Journal of Commerce (May 25, 2003) reported that Frontier Airlines conducted a study to estimate average passenger plus carry-on weights. They found an average summer weight of 183 lb and a winter average of 190 lb. Suppose that each of these estimates was based on a random sample of 100 passengers and that the sample standard deviations were 20 lb for the summer weights and 23 lb for the winter weights. A. Construct and interpret a 95% confidence interval for the mean summer weight (including carry-on luggage) of Frontier Airlines passengers.B. Construct and interpret a 95% confidence interval for the mean winter weight (including carry-on luggage) of Frontier Airlines passengers.C. The new FAA recommendations are 190 lb for sum- mer and 195 lb for winter. Comment on these recommen- dations in light of the confidence interval estimates from Parts (a) and (b).
Answer:
a) [tex]183-1.984\frac{20}{\sqrt{100}}=179.032[/tex]
[tex]183+1.984\frac{20}{\sqrt{100}}=186.968[/tex]
So on this case the 95% confidence interval would be given by (179.032;186.968)
b) [tex]190-1.984\frac{23}{\sqrt{100}}=185.437[/tex]
[tex]190+1.984\frac{23}{\sqrt{100}}=194.563[/tex]
So on this case the 95% confidence interval would be given by (185.437;194.563)
c) For Summer the confidence interval was (179.032;186.968) and as we can see our upper limit is <190 so then we can conclude that they are below the specification of 190 at 5% of significance
For Winter the confidence interval was (185.437;194.563) and again the upper limit is <190 so then we can conclude that they are below the specification of 195 at 5% of significance
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 : Summer
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)
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=100-1=99[/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(0.025,99)".And we see that [tex]t_{\alpha/2}=1.984[/tex]
Now we have everything in order to replace into formula (1):
[tex]183-1.984\frac{20}{\sqrt{100}}=179.032[/tex]
[tex]183+1.984\frac{20}{\sqrt{100}}=186.968[/tex]
So on this case the 95% confidence interval would be given by (179.032;186.968)
Part b: Winter
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)
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=100-1=99[/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(0.025,99)".And we see that [tex]t_{\alpha/2}=1.984[/tex]
Now we have everything in order to replace into formula (1):
[tex]190-1.984\frac{23}{\sqrt{100}}=185.437[/tex]
[tex]190+1.984\frac{23}{\sqrt{100}}=194.563[/tex]
So on this case the 95% confidence interval would be given by (185.437;194.563)
Part c
For Summer the confidence interval was (179.032;186.968) and as we can see our upper limit is <190 so then we can conclude that they are below the specification of 190 at 5% of significance
For Winter the confidence interval was (185.437;194.563) and again the upper limit is <190 so then we can conclude that they are below the specification of 195 at 5% of significance
Final answer:
Using statistical methods, we have constructed 95% confidence intervals for the mean summer and winter weights of Frontier Airlines passengers, finding that the new FAA recommendations align with our calculated intervals but are positioned conservatively towards the upper limits.
Explanation:
To answer the student's question about constructing and interpreting 95% confidence intervals for the mean summer and winter weights of Frontier Airlines passengers, and then commenting on the new FAA recommendations, we'll follow statistical methods.
Part A: Summer Weight
For the summer weight, the mean is 183 lb, the standard deviation is 20 lb, and the sample size (n) is 100 passengers. The formula for a 95% confidence interval is: mean ± (z*standard error), where the standard error is (standard deviation/√n). Using z=1.96 for 95% confidence, we calculate the interval as 183 ± (1.96*(20/√100)), which simplifies to 183 ± 3.92. Thus, the 95% confidence interval for summer is ±183 lb ± 3.92 lb.
Part B: Winter Weight
Similar calculations for the winter weight (mean = 190 lb, standard deviation = 23 lb, n = 100) yield a 95% confidence interval of 190 ± (1.96*(23/√100)), or 190 ± 4.51. Therefore, the 95% confidence interval for winter is ±190 lb ± 4.51 lb.
Part C: Comment on FAA Recommendations
The new FAA recommendations are 190 lb for summer and 195 lb for winter. These fall within our confidence intervals, but lean towards the upper limit, suggesting a conservative approach by the FAA, possibly to ensure safety by accommodating potential underestimations of passenger weights.
The parents of three children, ages 1, 3, and 6, wish to set up a trust fund that will pay X to each child upon attainment of age 18, and Y to each child upon attainment of age 21. They will establish the trust fund with a single investment of Z. Find the equation of value for Z. ?
To calculate the initial trust fund deposit for three children of diverse ages, you create an equation reflecting the present value of the future payouts for each child at ages 18 and 21. The equation of value here models the required single investment with respect to the annual interest rate.
Explanation:The problem presented, concerning setting up a trust fund for three children aged 1, 3, and 6, involves future value of lump sum investments, compound interest, and time value of money. Using an equation of value approach, the equation for the single investment Z made today that will pay each child X at age 18 and Y at age 21 can be modeled as follows:
[tex]Z= (X/(1+r)^{(18-1)} + Y/(1+r)^{(21-1)}) + (X/(1+r)^{(18-3)} + Y/(1+r)^{(21-3)} ) + (X/(1+r)^{(18-6)} + Y/(1+r)^{(21-6)})[/tex]where r represents the annual interest rate. This equation reflects the present value of the future payouts for each child. The term (1+r)years until payout is used to calculate the present value of each future payment. The equation adds up these present values for each child to calculate the initial trust fund deposit (Z).
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Explain the difference between the interquartile range and the range. Which is more sensitive to extreme values? Explain your thinking.
Answer:
Interquartile range: Difference between the third and the first quartile.
Range: Difference between the highest and the lowest value of a dataset.
Since the range is not in the middle of a set(like the quartiles), it is more sensitive to extreme values.
Step-by-step explanation:
The interquartile range is the difference between the third quartile and the first quartile.
The first quartile(Q1) separates the lower 25% from the upper 75% of a set. So 25% of the values in a data set lie at or below the first quartile, and 75% of the values in a data set lie at or above the first quartile.
The third quartile(Q3) separates the lower 75% from the upper 25% of a set. So 75% of the values in a data set lie at or below the third quartile, and 25% of the values in a data set lie at or the third quartile.
The range is the difference between the highest and the lowest value of a dataset.
Lets see two different sets of cardinality 8
Set 1: 5,5,6,7,8,9,13,13
Range = 13-5 = 8
Q1 = Element at the position 0.25*8 = 2 = 5
Q3 = Element at the position 0.75*8 = 6 = 9
IQR = Q3 - Q1 = 9 - 5 = 4
Now, replacing the first and the last element by extreme values
Set 1: 0,5,6,7,8,9,13,20
Range = 20-0 = 20
Q1 = Element at the position 0.25*8 = 2 = 5
Q3 = Element at the position 0.75*8 = 6 = 9
IQR = Q3 - Q1 = 9 - 5 = 4
The range changes and the IQR does not. Since extreme values are not in the middle of the distribution(in which the quartiles are), the range is more sensitive to extreme values, as this example showed.
Final answer:
The interquartile range measures the spread of the middle 50% of data and is less sensitive to extreme values compared to the range, which measures the spread of all data points by looking at the difference between the maximum and minimum values.
Explanation:
Difference Between Interquartile Range and Range in Data Analysis:
The difference between the interquartile range (IQR) and the range is that while the range measures the spread of all data points by subtracting the smallest value from the largest, the interquartile range measures the spread of the middle 50% of data, between the 25th and 75th percentiles (Q1 and Q3 respectively). To calculate the range in a dataset, if the highest score is 100 and the lowest is 70, the range is 100 - 70 = 30. On the other hand, if Q1 is 72.25 and Q3 is 86.75, the interquartile range would be 86.75 - 72.25 = 14.50.
The range is more sensitive to extreme values or outliers because it takes into account the most extreme scores in the dataset. In contrast, the IQR is less sensitive to extreme values since it focuses on the central portion of the data set and ignores the extremes. This is why IQR is often used to identify potential outliers, which are suspected if they lie more than 1.5 times the IQR below Q1 or above Q3.
Therefore, the IQR provides a more robust measure of data variability, especially when dealing with skewed distributions or datasets with outliers.
You go to the Huron Valley Humane Society so you can adopt a dog.
For each random variable below, determine whether it is categorical, quantitative discrete, or quantitative continuous.
a) The number of days (to the nearest day) the dog has been at Huron Valley Humane Society
b) Whether or not the dog has a microchip
c) The breed of the dog
d) How much the dog weighs (in pounds)
e) The amount of food (in cups) the dog eats
f) The number of people who have taken the dog out for a walk
g) Whether you decide to adopt the dog
Answer:
a) quantitative discrete data
b) categorical variable
c) categorical variable
d) quantitative continuous variable
e) quantitative discrete variable
f) categorical variable
Step-by-step explanation:
Categorical variable are the non parametric variables. Their value cannot be expressed in the form of numerical.Quantitative discrete are the variables whose values are expressed in whole numbers. These variable cannot take decimal values and hence cannot take all the values within interval.Quantitative continuous variable values can be expressed in the form of decimals and they can take any value within an interval.a) The number of days (to the nearest day) the dog has been at Huron Valley Humane Society
Since days will always have whole number numerical values it is a quantitative discrete data.
b) Whether or not the dog has a microchip
This will be answered with a yes or a no. Thus, it is a categorical variable. It does not have any numerical value.
c) The breed of the dog
This is a categorical variable. It does not have any numerical value. It will have non-parametric values.
d) How much the dog weighs (in pounds)
Since weight is always measured and not counted. Also weight can take decimal values, thus it is quantitative continuous variable.
e) The amount of food (in cups) the dog eats
Since this will have whole number values. Also, number of cups will be counted. Thus, it is a quantitative discrete variable.
f) The number of people who have taken the dog out for a walk
Since this will have whole number values. Also, number of people will be counted. Thus, it is a quantitative discrete variable.
g) Whether you decide to adopt the dog
This will be answered with a yes or a no. Thus, it is a categorical variable. It does not have any numerical value.
a,f are quantitave discrete b,c,g are categorical d,e,are quantitative continous.
a.he number of days (to the nearest day) the dog has been at Huron Valley Humane Society: This is quantitative discrete data because the number of days is counted in whole numbers.
b.Whether or not the dog has a microchip: This is categorical data because it categorizes the dog as either having or not having a microchip.
c.The breed of the dog: This is categorical data because breeds represent different categories.
d.How much the dog weighs (in pounds): This is quantitative continuous data because weight can be measured to a very fine degree.
e.The amount of food (in cups) the dog eats: This is quantitative continuous data because the amount can be measured to any level of precision.
f.The number of people who have taken the dog out for a walk: This is quantitative discrete data because it involves counting distinct individuals.
g.Whether you decide to adopt the dog: This is categorical data because it categorizes your decision into adopt or not adopt.
A statistics professor finds that when he schedules an office hour for student help, an average of 3.3 students arrive. Find the probability that in a randomly selected office hour, the number of student arrivals is 6.
Answer:
[tex] P(X=6)[/tex]
If we use the probability mass function we got:
[tex] P(X=6) = \frac{e^{-3.3} 3.3^6}{6!}= 0.0662[/tex]
Step-by-step explanation:
Previous concepts
The Poisson process is useful when we want to analyze the probability of ocurrence of an event in a time specified. The probability distribution for a random variable X following the Poisson distribution is given by:
Solution to the problem
Let X the random variable that represent the number of students arrive at the office hour. We know that [tex]X \sim Poisson(\lambda=3.3)[/tex]
The probability mass function for the random variable is given by:
[tex]f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...[/tex]
And f(x)=0 for other case.
For this distribution the expected value is the same parameter [tex]\lambda[/tex]
[tex]E(X)=\mu =\lambda=3.3[/tex]
And we want this probability:
[tex] P(X=6)[/tex]
If we use the probability mass function we got:
[tex] P(X=6) = \frac{e^{-3.3} 3.3^6}{6!}= 0.0662[/tex]
Answer:
0.0662
Step-by-step explanation:
The given problem indicates the Poisson experiment.
The pdf of Poisson experiment is as follow
P(X=x)=μ^x(e^-μ)/x!
Here, the average student visits in office hour=3.3=μ.
We have to find the probability that the number of student arrivals is 6 in a randomly selected office hour.
So, x=6 and μ=3.3
The probability that the number of student arrivals is 6 in a randomly selected office hour
P(X=6)=3.3^6(e^-3.3)/6!
P(X=6)=1291.468(.0369)/720
P(X=6)=47.6552/720
P(X=6)=0.0662.
Thus, the probability that the number of student arrivals is 6 in a randomly selected office hour is 6.62 or 0.0662.
The heights, in inches, of the starting five players on a college basketball team are 6868, 7373, 7777, 7575, and 8484. Considering the players as a sample, the mean and standard deviation of the heights are 75.475.4 inches and 5.95.9 inches, respectively. When the players are regarded as a population, the mean and standard deviation of the heights are 75.475.4 inches and 5.25.2 inches, respectively. Explain why, numerically, the sample mean of 75.475.4 inches is the same as the population mean but the sample standard deviation of 5.95.9 inches differs from the population standard deviation of 5.25.2 inches
Answer:
The sample standard deviation of 5.95.9 inches differs from the population standard deviation of 5.25.2 inches because of their formulas for calculating it.
Step-by-step explanation:
We are given the heights, in inches, of the starting five players on a college basketball team ;
68, 73, 77, 75 and 84
Now whether we treat this data as sample data or population data, the mean height would remain same in both case because the formula for calculating mean is given by ;
Mean = Sum of all data values ÷ No. of observations
Mean = ( 68 + 73 + 77 + 75 + 84 ) ÷ 5 = 75.4 inches
So, numerically, the sample mean of 75.4 inches is the same as the population mean.
Now, coming to standard deviation there will be difference in both sample and population standard deviation and that difference occurs due to their formulas;
Formula for sample standard deviation = [tex]\frac{\sum (X_i - Xbar)^{2} }{n-1}[/tex]
where, [tex]X_i[/tex] = each data value
X bar = Mean of data
n = no. of observations
Sample standard deviation = [tex]\frac{ (68 - 75.4)^{2} +(73 - 75.4)^{2}+(77- 75.4)^{2}+(75- 75.4)^{2}+(84 - 75.4)^{2} }{5-1}[/tex]
= 5.9 inches
Whereas, Population standard deviation = [tex]\frac{\sum (X_i - Xbar)^{2} }{n}[/tex]
= [tex]\frac{ (68 - 75.4)^{2} +(73 - 75.4)^{2}+(77- 75.4)^{2}+(75- 75.4)^{2}+(84 - 75.4)^{2} }{5}[/tex] = 5.2 inches .
So, that's why sample standard deviation of 5.95.9 inches differs from the population standard deviation of 5.25.2 inches only because of formula.
In 1998 the World Health Organization reported the findings of a major study on the quality of blood pressure monitoring around the world. In its report it stated that for Canada the results for diastolic blood pressure had a mean of 78 mmHg and a standard deviation of 11 mmHg. Assuming that diastolic blood pressure measurements are Normally distributed, the DBP reading that represents the 80th percentile of the distribution is _____.
Answer:
87.26 mm Hg is the reading that represents the 80th percentile of the distribution.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 78 mm Hg
Standard Deviation, σ = 11 mm Hg
We are given that the distribution of diastolic blood pressure 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 value of x such that the probability is 0.80
P(X < x)
[tex]P( X < x) = P( z < \displaystyle\frac{x - 78}{11})=0.8[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(z < 0.842) = 0.8[/tex]
[tex]\displaystyle\dfrac{x - 78}{11} = 0.842\\\\x = 87.262 \approx 87.26[/tex]
87.26 mm Hg is the reading that represents the 80th percentile of the distribution.
To find the DBP reading representing the 80th percentile, we can use the z-score formula. The DBP reading is 86.8 mmHg.
Explanation:To find the diastolic blood pressure (DBP) reading that represents the 80th percentile of the distribution, we can use the z-score formula. The z-score is calculated by subtracting the mean from the desired value and dividing it by the standard deviation. Then, we can use the z-score table or calculator to find the corresponding percentile. In this case, since we want the 80th percentile, we need to find the z-score that corresponds to an area of 0.8 to the left of it.
The formula for the z-score is: z = (x - mean) / standard deviation
Using the given values for Canada (mean = 78 mmHg, standard deviation = 11 mmHg), we can substitute them into the formula and solve for x:
0.8 = (x - 78) / 11
Multiplying both sides by 11, we get:
8.8 = x - 78
Adding 78 to both sides, we get:
x = 86.8 mmHg
Therefore, the DBP reading that represents the 80th percentile of the distribution is 86.8 mmHg.
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chang drank 10 fluid ounces of juice. how much is this in cups? write your answer as a whole number or a mixed number in simplest form.
Answer:
1 1/4 cups
Step-by-step explanation:
Their are 8 ounces in one cup and their are 2 ounces in 1/4's of a cup
CUP FL. OZ
1 8
3/4 6
2/3 5
1/2 4
1/3 3
1/4 2
1/8 1
1/16 0.5
Suppose that 30% of all students who have to buy a text for a particular course want a new copy (the successes!), whereas the other 70% want a used copy. Consider randomly selecting 15 purchasers.
(a) What are the mean value and standard deviation of the number who want a new copy of the book?(b) What is the probability that the number who want new copies is more than two standard deviations away from the mean value?
Answer:
a. Mean = 4.5, Standard Deviation = 1.775
b. 0.0152
Step-by-step explanation:
Given
n = 15 purchasers
p = Success = 30%
p = 0.3
q =Failure = 70%
q = 0.7
a.
Mean = np
Mean = 15 * 0.3
Mean = 4.5
Standard Deviation = √Variance
Variance = npq
Variance = 15 * 0.3 * 0.7
Variance = 3.15
Standard Deviation = √3.15
Standard Deviation = 1.774823934929884
Standard Deviation = 1.775 ---------- Approximated
b.
The probability that the number who want new copies is more than two standard deviations away from the mean value
Standard Deviation = 1.775
Mean = 4.5
2 Standard Deviation and Mean = 2 * 1.775 + 4.5
= 3.55 + 4.5
= 8.05
P(X>8.05) = P(9) + P(10) +........+ P(15)
Using the binomial distribution
(p + q) ^ n where p = 0.3, q = 0.7 , n = 15
Expanding (p+q)^n where n = 15 and r > 8
We have
15C9 p^9 q^6 + 15C10 p^10 q^5 + 15C11 p^11 q⁴ + 15C12 p^12 q³ + 15C13 p^13 q² + 15C14 p^14 q + p^15
= 5005 (0.3)^9 (0.7)^6 + 3003 (0.3)^10 (0.7)^5 + 1365 (0.3)^11 (0.7)⁴ + 455 (0.3)^12 (0.7)³ + 105 (0.3)^14 (0.7)² + 15 (0.3)^14 (0.7) + 0.3^15
=0.015234
a marketing survey compiled data on the total number of televisions in households where k is a positie constant. What is the probability that a randomly chosen household has at least two televisions?
The probability that a randomly chosen household has at least two televisions can be calculated by dividing the number of households with at least two televisions by the total number of households.
Explanation:To find the probability that a randomly chosen household has at least two televisions, we need to use the data from the marketing survey. Let's assume there are 'n' households in total. The probability of a household having at least two televisions can be calculated by dividing the number of households with at least two televisions by the total number of households, which is 'n'.
Let's say there are 'x' households with at least two televisions. The probability can be expressed as:
P(at least 2 TVs) = x/n
For example, if there are 100 households in total and 30 of them have at least two televisions, then the probability would be P(at least 2 TVs) = 30/100 = 0.3, which is 30%.
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At least 96.00% of the data in any data set lie within how many standard deviations of the mean? Explain how you arrived at your answer.
At least 96.00% of the data in any data set lies within three standard deviations of the mean.
Explanation:The Empirical Rule states that in a bell-shaped and symmetric distribution, approximately 68 percent of the data lies within one standard deviation of the mean, 95 percent lies within two standard deviations of the mean, and more than 99 percent lies within three standard deviations of the mean. Therefore, at least 96.00% of the data lies within three standard deviations of the mean.
Three students were applying to the same graduate school. They came from schools with different grading systems. Student GPA School Average GPA School Standard Deviation Thuy 2.7 3.2 0.8 Vichet 88 75 20 Kamala 8.8 8 0.4 Which student had the best GPA when compared to other students at his school? Explain how you determined your answer. (Enter your standard deviation to two decimal places.) had the best GPA compared to other students at his school, since his GPA is standard deviations
Answer:
Kamala had the best GPA compared to other students at his school, since his GPA is 2 standard deviations above his school's mean.
Step-by-step explanation:
The z-score measures how many standard deviation a score X is above or below the mean.
it is given by the following formula:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
In which
[tex]\mu[/tex] is the mean, [tex]\sigma[/tex] is the standard deviation.
In this problem:
The student with the best GPA compared to other students at his school is the one with the higher z-score, that is, the one whose grade is the most standard deviations above the mean for his school
Thuy
Student GPA| School Average GPA| School Standard Deviation
Thuy 2.7| 3.2| 0.8
So [tex]X = 2.7, \mu = 3.2, \sigma = 0.8[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{2.7 - 3.2}{0.8}[/tex]
[tex]Z = -0.625[/tex]
Thuy's score is -0.625 standard deviations below his school mean.
Vichet
Student GPA| School Average GPA| School Standard Deviation
Vichet 88| 75| 20
So [tex]X = 88, \mu = 75, \sigma = 20[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{88 - 75}{20}[/tex]
[tex]Z = 0.65[/tex]
Vichet's score is 0.65 standard deviation above his school mean
Kamala
Student GPA| School Average GPA| School Standard Deviation
Kamala 8.8| 8| 0.4
So [tex]X = 8.8, \mu = 8, \sigma = 0.4[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{8.8 - 8}{0.4}[/tex]
[tex]Z = 2[/tex]
Kamala's score is 2 standard deviations above his school mean.
Kamala has the higher z-score, so he had the best GPA when compared to other students at his school.
The correct answer is:
Kamala had the best GPA compared to other students at his school, since his GPA is 2 standard deviations above his school's mean.
Final answer:
Kamala had the best GPA relative to her school's average, with a z-score of 2.00, indicating her GPA is 2 standard deviations above the average.
Explanation:
To determine which student had the best GPA compared to other students at their school, we need to calculate how many standard deviations each student's GPA is away from their school's average GPA. This can be done using the formula:
z = (x - μ) / σ
Where z is the number of standard deviations, x is the student's GPA, μ (mu) is the school's average GPA, and σ (sigma) is the school's standard deviation.
Kamala has the highest z-score, which means her GPA is the furthest above her school's average when compared to Thuy and Vichet. Therefore, Kamala had the best GPA relative to her peers at her school.
What is the standard deviation of a random variable q with the following probability distribution? (Do not round intermediate calculations. Enter your answer in numbers not in percentage. Round your answer to 4 decimal places.) Value of q Probability 0 0.25 1 0.25 2 0.50
Answer:
The standard deviation of given probability distribution is 0.8292
Step-by-step explanation:
We are given the following in the question:
q: 0 1 2
Probability: 0.25 0.25 0.50
Formula:
[tex]E(q) = \displaystyle\sum q_ip(q_i)\\=0(0.25) + 1(0.25) + 2(0.50) = 1.25[/tex]
[tex]E(q^2)= \displaystyle\sum q_i^2p(q_i)\\=0^2(0.25) + 1^2(0.25) + 2^2(0.50) = 2.25[/tex]
Variance =
[tex]\sigma^2 = E(q^2) = (E(q))^2\\= 2.25- (1.25)^2\\=0.6875\\\sigma = \sqrt{0.6875} = 0.8292[/tex]
Thus, the standard deviation of given probability distribution is 0.8292
The standard deviation of a random variable can be calculated using the formula: √∑(xi-μ)2 ⋅ P(xi) . We find the mean (μ) by multiplying each value of q by its corresponding probability and summing them up. Next, we calculate the squared difference between each value of q and the mean, multiplied by their respective probabilities. Finally, we take the square root of this value to obtain the standard deviation.
Explanation:The standard deviation of a random variable can be calculated using the formula:
\sqrt{\sum (x_i - \mu)^2 \cdot P(x_i)}
First, we need to find the mean (μ) of the probability distribution. The mean can be calculated by multiplying each value of q by its corresponding probability and summing them up:
μ = (0 × 0.25) + (1 × 0.25) + (2 × 0.5) = 0.5
Next, we calculate the squared difference between each value of q and the mean, multiplied by their respective probabilities:
(0 - 0.5)^2 × 0.25 + (1 - 0.5)^2 × 0.25 + (2 - 0.5)^2 × 0.5 = 0.5
Finally, we take the square root of this value to obtain the standard deviation:
√0.5 ≈ 0.7071
Therefore, the standard deviation of the random variable q is approximately 0.7071.
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