At a large bank, account balances are normally distributed with a mean of $1,637.52 and a standard deviation of $623.16. What is the probability that a simple random sample of 400 accounts has a mean that exceeds $1,650?

Answers

Answer 1
Answer:

Answer:

P(\bar X >1650)=P(Z>(1650-1637.52)/((623.16)/(√(400)))=0.401)

And we can use the complement rule and we got:

P(Z>0.401) =1-P(Z<0.401) = 1-0.656= 0.344

Step-by-step explanation:

Previous concepts

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

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

Solution to the problem

Let X the random variable that represent the bank account balances of a population, and for this case we know the distribution for X is given by:

X \sim N(1637.52,623.16)  

Where \mu=1637.52 and \sigma=623.16

Since the distribution of X is normal then the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, (\sigma)/(√(n)))

And we can use the z score formula given by:

z = (\bar X -\mu)/((\sigma)/(√(n)))

And using this formula we got:

P(\bar X >1650)=P(Z>(1650-1637.52)/((623.16)/(√(400)))=0.401)

And we can use the complement rule and we got:

P(Z>0.401) =1-P(Z<0.401) = 1-0.656= 0.344

Answer 2
Answer:

Answer: the probability is 0.49

Step-by-step explanation:

Since the account balances at the large bank are normally distributed.

we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = account balances.

µ = mean account balance.

σ = standard deviation

From the information given,

µ = $1,637.52

σ = $623.16

We want to find the probability that a simple random sample of 400 accounts has a mean that exceeds $1,650. It is expressed as

P(x > 1650) = 1 - P(x ≤ 1650)

For x = 1650,

z = (1650 - 1637.52)/623.16 = 0.02

Looking at the normal distribution table, the probability corresponding to the z score is 0.51

P(x > 1650) = 1 - 0.51 = 0.49


Related Questions

Given the statement, TSR: ABC and the diagramof the right triangle TSR, determine theapproximate length of BC
Differentiate Functions of Other Bases In Exercise, find the derivative of the function.y = 3x
Mr Thompson wants to cut a 5-ft rope into a 1/4-ft sections how many 1/4 sections will he have
Identify any outliers in the data set. 35,41,44,45,47,58
What is the answer to this problem and why 60+60×0+1=

The length of a side of a triangle is 36. A line parallel to that side divides the triangle into two parts of equal area. Find the length of the segment determined by the points of intersection between the line and the other two sides of the triangle

Answers

Answer:

  18√2

Step-by-step explanation:

The area of the smaller triangle is 1/2 that of the larger one. Since the triangles are similar, the dimensions of the smaller triangle are √(1/2) those of the larger one.

  36 · √(1/2) = 36 · (√2)/2 = 18√2 . . . . length of line dividing the triangle

Suppose that B1 and B2 are mutually exclusive and complementary events, such that P(B1 ) = .6 and P(B2) = .4. Consider another event A such that P(A | B1) = .2 and P(A | B2) = .5. Find P(A).

Answers

Answer:

So, we get that is P(A)=0.32.

Step-by-step explanation:

We know that:

P(B_1)=0.6\n\nP(B_2)=0.4\n\nP(A|B_1)=0.2\n\nP(A|B_2)=0.5\n

We have the formula for probability:

P(A|B)=(P(A\cap B))/(P(B))\n\n\implies P(A\cap B)=P(A|B)\cdot P(B)

So, we calculate:

P(A\cap B_1)=P(A|B_1)\cdot P(B_1)\n\nP(A\cap B_1)=0.2\cdot 0.6=0.12\n\n\nP(A\cap B_2)=P(A|B_2)\cdot P(B_2)\n\nP(A\cap B_2)=0.5\cdot 0.4=0.2\n

We calculate:

P(A)=P((A\cap B_1)\cup(A\cap B_2))\n\nP(A)=P(A\cap B_1)+P(A\cap B_2)\n\nP(A)=0.12+0.2\n\nP(A)=0.32

So, we get that is P(A)=0.32.

Final answer:

To find P(A), use the law of total probability given that B1 and B2 are mutually exclusive and complementary events. Substituting the provided values, P(A) = 0.32.

Explanation:

The question is asking us to calculate P(A), given the values for P(A | B1) and P(A | B2), and the knowledge that B1 and B2 are mutually exclusive and complementary events. In probability, if events B1 and B2 are mutually exclusive and complementary, this means that one and only one of them can occur, and their occurrence covers all possible outcomes. We can use the law of total probability to find the overall P(A). The law of total probability states that P(A) = P(A | B1) * P(B1) + P(A | B2) * P(B2). Plugging the provided values into this formula, we get P(A) = .2 * .6 + .5 * .4 = .12 + .2 = .32. Therefore, P(A) is .32.

Learn more about Probability here:

brainly.com/question/22962752

#SPJ11

What is the number if 2 is 12.5% of a number?
PLS HELP ME!!!

Answers

16 since 12.5% is 1/8 of 100 and 2 times 8 is 16

The answer to the question is 25.

Write the partial fraction decomposition of the rational expression. Check your result algebraically.

Answers

Answer:

See below.

Step-by-step explanation:

First, distribute:

=(1)/(x(x+1))

Now, perform partial fraction decomposition. This is only two factors, so we only need linear functions:

(1)/(x(x+1)) =(A)/(x)+(B)/(x+1)

Now, multiply everything by x(x+1):

1=A(x+1)+B(x)

Now, solve for each variable. Let's let x=-1:

1=A(-1+1)+B(-1)

1=0A-B=-B

B=-1

Now, let's let x=0:

1=A(0+1)+B(0)

A=1

So:

(1)/(x(x+1))=(1)/(x)-(1)/((x+1))

Double Check:

(1)/(x)-(1)/((x+1))=((x+1))/(x(x+1))-(x)/(x(x+1))

=(x-x+1)/(x(x+1))    =(1)/(x^2+x)

a group of students is donating blood during a blood drive. a student has 9/20 probability of having type 'o' blood and a 2/5 probability of having type 'A'.what is the probability that a student has type 'o' or type 'a' blood and why?

Answers

For the answer to the question above,
the answer is "The student can have only one blood type, so the actual events are mutually exclusive. "

The probabilities are not mutually exclusive. Based on the group 

P(Type O) = 9/20 = 45% or 0.45 
P(Type A) = 2/5 = 40% or 0.40 
P(Other) = 3/20 = 15% or 0.15
I hope my answer helped you. Feel free to ask more questions. Have a nice day!
Hello there.

A group of students is donating blood during a blood drive. a student has 9/20 probability of having type 'o' blood and a 2/5 probability of having type 'A'.what is the probability that a student has type 'o' or type 'a' blood and why?

The student can have only one blood type, so the actual events are mutually exclusive.

The upcoming championship high school football game is a big deal in your little town. The problem is, it is being played in the next biggest town, which is two hours away! To get as many people as you can to attend the game, you decide to come up with a ride-sharing app, but you want to be sure it will be used before you put all the time in to creating it. You determine that if more than three students share a ride, on average, you will create the app.You conduct simple random sampling of 20 students in a school with a population of 300 students to determine how many students are in each ride-share (carpool) on the way to school every day to get a good idea of who would use the app. The following data are collected:6 5 5 5 3 2 3 6 2 25 4 3 3 4 2 5 3 4 5Construct a 95% confidence interval for the mean number of students who share a ride to school, and interpret the results.Part A: State the parameter and check the conditions.Part B: Construct the confidence interval. Be sure to show all your work, including the degrees of freedom, critical value, sample statistics, and an explanation of your process.Part C: Interpret the meaning of the confidence interval.Part D: Use your findings to explain whether you should develop the ride-share app for the football game.

Answers

The parameter used in the probability is the average number of students represented by u.

How to calculate the probability?

The confidence interval based on the information will be:

= 3.85 - 2.09(1.348 / ✓20)

= 3.22

Also, 3.85 + 2.09(1.348 / ✓20) = 4.48

The confidence interval simply means that one is 95% confident that the true mean is between 3.22 and 4.48.

Learn more about probability on:

brainly.com/question/24756209

Answer:

a) The parameter of interest on this case is \mu who represent the average number of students in order to create an app.

b) 3.85 - 2.09 (1.348)/(√(20))=3.22  

3.85 + 2.09 (1.348)/(√(20))=4.48  

The 95% confidence interval is given by (3.22;4.48)  

c) For this case we have 95% of confidence that the true mean for the average number of students in order to create an app is between 3,22 and 4.48.

d) We have the following criteria in order to decide: "You determine that if more than three students share a ride, on average, you will create the app"

So then since the lower limti for the confidence interval is higher than 3 we can conclude that at 5% of significance we have more than 3 students share a ride so then makes sense create the app.

Step-by-step explanation:

Part a

The parameter of interest on this case is \mu who represent the average number of students in order to create an app.

Part b

Data: 6 5 5 5 3 2 3 6 2 2 5 4 3 3 4 2 5 3 4 5

n=20 represent the sample size  

\bar X represent the sample mean  

s represent the sample standard deviation  

m represent the margin of error  

Confidence =95% or 0.95

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".  

Calculate the mean and standard deviation for the sample

On this case we need to find the sample standard deviation with the following formula:

s=sqrt{(\sum_(i=1)^20 (x_i -\bar x)^2)/(n-1)}

And in order to find the sample mean we just need to use this formula:

\bar x =(\sum_(i=1)^(20) x_i)/(n)

The sample mean obtained on this case is \bar x= 3.85 and the deviation s=1.348

Calculate the critical value tc  

In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. The degrees of freedom are given by:  

df=n-1=20-1=19  

We can find the critical values in excel using the following formulas:  

"=T.INV(0.025,19)" for t_(\alpha/2)=-2.09  

"=T.INV(1-0.025,19)" for t_(1-\alpha/2)=2.09  

The critical value tc=\pm 2.09  

Calculate the confidence interval  

The interval for the mean is given by this formula:  

\bar X \pm t_(c) (s)/(√(n))  

And calculating the limits we got:  

3.85 - 2.09 (1.348)/(√(20))=3.22  

3.85 + 2.09 (1.348)/(√(20))=4.48  

The 95% confidence interval is given by (3.22;4.48)  

Part c

For this case we have 95% of confidence that the true mean for the average number of students in order to create an app is between 3.22 and 4.48.

Part d

We have the following criteria in order to decide: "You determine that if more than three students share a ride, on average, you will create the app"

So then since the lower limti for the confidence interval is higher than 3 we can conclude that at 5% of significance we have more than 3 students share a ride so then makes sense create the app.