An airplane with room for 100 passengers has a total baggage limit of 6000 lb. Suppose that the total weight of the baggage checked by an individual passenger is a random variable x with a mean value of 49 lb and a standard deviation of 18 lb. If 100 passengers will board a flight, what is the approximate probability that the total weight of their baggage will exceed the limit? (Hint: With n = 100, the total weight exceeds the limit when the average weight x exceeds 6000/100.) (Round your answer to four decimal places.)

Answer:

0.706

Step-by-step explanation:  

For uniformity, let's assuming none of the people selected is born on 29th of February. Therefore, the total possible birthday (sample space) for anybody selected is 365 days.  

Probability of having at least two people having or not having the same birthday sum up to 1. It is easier to calculate the probability of people not having the same birthday.  

The probability of at least two people having the same birthday = 1 - The probability of at least two people not having the same birthday  

Pr (First Person selected is born in a day in a year) = [tex]\frac{365}{365}[/tex]  

The next person selected is limited to 364 possible days

Pr (Second Person selected is born in a day in year) = [tex]\frac{364}{365}[/tex]  

The next person selected is limited to 363 possible days

Pr (Third Person selected is born in a day in year) = [tex]\frac{363}{365}[/tex]

The next person selected is limited to 362 possible days

Pr (Fourth Person selected is born in a day in year) = [tex]\frac{362}{365}[/tex]

The next person selected is limited to 361 possible days

Pr (Fifth Person selected is born in a day in year) = [tex]\frac{361}{365}[/tex]  

The next person selected is limited to 360 possible days

Pr (Sixth Person selected is born in a day in year) = [tex]\frac{360}{365}[/tex]

The next person selected is limited to 359 possible days

Pr (Seventh Person selected is born in a day in year) = [tex]\frac{359}{365}[/tex]  

Looking at the pattern, the pattern continues by descending by 1 day from 365. The last person selected will have 336 possible days (365 – 29 days ) the other people before have used up 29 potential days.  

Pr (Last Person selected is born in a day in year) = [tex]\frac{359}{365}[/tex]  

The probability of selecting at least two people that will not have the same birthday, is the product of their 30 individual birthday probability.

[tex]\frac{365}{365} * \frac{364}{365} * \frac{363}{365} * \frac{362}{365} * \frac{361}{365} * \frac{360}{365} * \frac{359}{365} ... \frac{337}{365} * \frac{336}{365} = [/tex]  

This is a huge a hug e calculation to simplify. For simplicity the numerator and the denominator will be calculated separately.

The denominator is product of 365 in 30 times.  

The denominator = 365³⁰

To simplify the numerator, the factorials method is used. Using factorials method, 365! = 365 × 364 × 363 × 362 × 361 × 360 × 359 × 358 ... × 3 × 2 × 1. But we only need the product of the integers from 365 to 336, so we’ll divide the unwanted numbers by dividing 365! by 335!  

The numerator = [tex]\frac{365!}{335!}[/tex]

Therefore,

                       [tex]\frac{365!}{335! * 365^{30}} = 0.294[/tex]

The probability that no one selected in the group has the same birthday is 0.294. The probability that at least two people will have the same birthday is the complement of the probability that no one in the group has the same birthday.

                                       1 - 0.294 = 0.706

The probability that at least two of them will have the same birthday = 0.706

The correct answer is approximately 0.7064.

To solve this problem, it is easier to calculate the probability that no one shares a birthday and then subtract this from 1 to find the probability that at least two people share a birthday.

Let's denote the probability that no two people share a birthday as P(no shared birthday). There are 365 days in a year, so the first person can have any birthday.

The second person then has 364 out of 365 days to have a different birthday, the third person has 363 out of 365 days, and so on, until the thirtieth person has 336 out of 365 days to avoid sharing a birthday with anyone else.

Thus, the probability that no one shares a birthday is:

[tex]\[ P(\text{no shared birthday}) = \frac{365}{365} \times \frac{364}{365} \times \frac{363}{365} \times \ldots \times \frac{336}{365} \][/tex]

We can calculate this product to find P(no shared birthday). Then, to find the probability that at least two people share a birthday, we subtract this from 1:

[tex]\[ P(\text{at least one shared birthday}) = 1 - P(\text{no shared birthday}) \][/tex]

Calculating the product for P(no shared birthday) and subtracting from 1 gives us the probability that at least two people in the group of thirty will have the same birthday.

When rounded to four decimal places, this probability is approximately 0.7064.

Answer:   (1.1838, 1.1962)

Step-by-step explanation:

The formula we use to calculate the confidence interval for population mean ( if population standard deviation is not given)is given by :-

[tex]\overline{x}\pm t*\dfrac{s}{\sqrt{n}}[/tex],

where n= sample size

s= sample standard deviation.

[tex]\overline{x}[/tex]= sample mean

t* = Two-tailed critical t-value.

Given : n= 25

Degree of freedom : df = n-1 =24

Significance level [tex]=\alpha=1-0.95=0.95[/tex]

Now from students' t-distribution table , check the t-value for significance level [tex]\alpha/2=0.025[/tex] and df=24:

t*=2.0639

[tex]\overline{x}=1.19[/tex]  fluid ounces

s= 0.015  fluid ounces

We assume that the population is approximately normally distributed

Now, the 95% two-sided confidence interval on the mean volume of syrup dispensed :-

[tex]1.19\pm (2.0639)\dfrac{0.015}{\sqrt{25}}\\\\=1.19\pm (2.0639)(\dfrac{0.015}{5})\\\\=1.19\pm0.007728=(1.19-0.0061917,\ 1.19+0.0061917)\\\\=(1.1838083,\ 1.1961917)\approx(1.1838,\ 1.1962)[/tex]

∴ The required confidence interval = (1.1838, 1.1962)

To find the 95% two-sided confidence interval for the mean volume of syrup dispensed, use the t-distribution with the given sample mean, standard deviation, and size. The resulting interval is (1.183808, 1.196192) fluid ounces.

To calculate the 95% two-sided confidence interval for the mean volume of syrup dispensed, we can use the t-distribution, as the population standard deviation is unknown, and the sample size is less than 30.

Step-by-Step Explanation:

First, identify the sample mean (ar{x}), which is 1.19 fluid ounces, the sample standard deviation (s), which is 0.015 fluid ounces, and the sample size (n), which is 25.

Next, find the t-score that corresponds to a 95% confidence level and 24 degrees of freedom (n-1). You can use a t-distribution table or calculator for this, which typically gives you a t-score around 2.064.

Now calculate the margin of error (ME) using the formula ME = t*(s/sqrt{n}).

Finally, construct the confidence interval by adding and subtracting the margin of error from the sample mean: (1.19 - ME, 1.19 + ME).

Plugging in the values, we get ME = 2.064*(0.015/sqrt{25}) = 2.064*0.003 = 0.006192. So, the 95% confidence interval is (1.19 - 0.006192, 1.19 + 0.006192) = (1.183808, 1.196192).

Answer:

$12.50

Step-by-step explanation:

so Juan got his sweater at a 25% discount. So we woukd find 25% of 10 and then add it to $10

Answer:

Step-by-step explanation:

Juan only paid $10 for his new Sweater. He got a 25% discount

Let the original price of the sweater be $x. If he gets s 35% discount, it means that the price which she paid was lower than the original price.

Discount of 25% on the original price = 25/100 × x = 0.25x

The amount of money that Juan paid is the original price minus the discount of 25%. It becomes

x - 0.25x = 0.75x

Amount paid = $10. Therefore

0.75x = 10

x = 10/0.75 = $13.33

Answer:

x=11

Step-by-step explanation:

Answer:

(a) and (b) see pictures attached

(c) V = 16/35

Step-by-step explanation:

(a) Sketch the base of S in the xy-plane.

See picture 1 attached

(b) Sketch a three-dimensional picture of S with the xy-plane as the floor.

See picture 2 attached

(c) Compute the volume of S.

The volume is given by the triple integral

[tex]\displaystyle\iiint_{S}zdzdydx[/tex]

The cross-sections perpendicular to the x-axis are squares so  

[tex]z=1-x^2[/tex]

The region S is given by the following inequalities

[tex]0\leq x\leq 1\\\\x^2\leq y\leq 1\\\\0\leq z\leq 1-x^2[/tex]

Therefore

[tex]\displaystyle\iiint_{S}zdzdydx=\displaystyle\int_{0}^{1} \displaystyle\int_{x^2}^{1} \displaystyle\int_{0}^{1-x^2} (1-x^2)dzdydx=\\\\\displaystyle\int_{0}^{1}(1-x^2)(1-x^2)(1-x^2)dx=\displaystyle\int_{0}^{1}(1-x^2)^3dx=\displaystyle\frac{16}{35}[/tex]

So the volume V of the solid S is

V=16/35

Answer:

f(x) =4x² + x - 7

Step-by-step explanation:

f(x) = ax² + bx + c

f(1) = a + b + c

f(-3) = 9a - 3b + c

f(3) = 9a + 3b + c

a + b + c = -2 -----(1)

c = -2-a-b -----(2)

9a - 3b + c = 26 ------(3)

9a + 3b + c = 32 -------(4)

(2)->(3)

9a-3b-2-a-b = 26

8a-4b = 28

4(2a-b) = 28

2a-b = 7 ----(5)

(2)->(4)

9a+3b-2-a-b = 32

8a+2b = 34

2(4a+b) = 34

4a+b = 17 ----(6)

(5)+(6)

6a = 24

a = 4

sub a=4 into(6)

4(4)+b = 17

b = 1

sub a=4,b=1 into (2)

c = -2-4-1

c = -7

To test if the proportion of college graduates who believe in evolution differs significantly from the proportion of individuals with some college education, we can use the z-test statistic.

To test if there is evidence that the proportion of college graduates who believe in evolution differs significantly from the proportion of individuals with some college education who believe in evolution, we can use the z-test statistic. The formula for the z-test statistic is z = (p1 - p2) / sqrt(pq((1/n1) + (1/n2))), where p1 and p2 are the proportions of college graduates and individuals with some college education who believe in evolution, n1 and n2 are the respective sample sizes, and q = 1 - p. Plugging in the given values, we have z = (0.459- 133/325 - 121/228) / sqrt(0.459 * (1-0.459) * ((1/325) + (1/228))). Calculating this expression will give us the z-test statistic.

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Answer:

12%

Step-by-step explanation:

The probability that the residents of a household do not own two cars and have an income over $50,000 a year is 100% subtracted by the probability of owning two cars and having an income over $50,000(P(2∩50+) and the probability of not having an income over $50,000 (P(50-))

[tex]P = 1 - P(2\cap 50+) - P(50-)\\P = 1 - (0.6*0.7) - (1-0.6)\\P=1-0.48-0.40\\P=0.12[/tex]

Therefore, the probability that the residents of a household do not own two cars and have an income over $50,000 a year is 12%

Final answer:

To find the probability that a household has an income over $50,000 and does not own two cars, multiply the probability of a household not owning two cars (20%) by the probability of having an income over $50,000 (60%), which equals 12%.

Explanation:

The question is asking us to find the probability that the residents of a household do not own two cars and have an income over $50,000 a year. We know that the probability a household with income over $50,000 owns two cars is 80%, so the probability that such a household does not own two cars is 1 - 0.80 = 0.20 or 20%. Since 60% of households have an income over $50,000, we can calculate the probability that a household has an income over $50,000 and does not own two cars by multiplying these two probabilities: 0.60 * 0.20 = 0.12 or 12%.

The probability that the average income of a 41-person sample would be less than $42,000, given a population with mean annual income of $43,000 and a standard deviation of $29,000, is 0.4129.

This is a problem of probability concerning the Mean and Standard Deviation of the sampling distribution of the sample mean. We first have to find the Standard Error (SE) for the sampling distribution of the sample mean, which is given by SE = σ/√n , where σ is the standard deviation (29) and n is the sample size (41). Thus, SE = 29/√41 = 4.52.

Next, we find the Z score corresponding to the sample mean of 42. Z = (X - μ)/SE, where X is the sample mean (42), μ is the population mean (43), and SE is the standard error. So Z = (42 - 43)/4.52 = -0.22.

Looking up a Z score of -0.22 in a standard normal distribution table, we find a probability of 0.4129. Therefore, the answer is B) 0.4129.

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Final answer:

Using the Central Limit Theorem and provided statistics, the probability that the sample mean income is less than 42 thousand dollars is found by calculating a z-score and consulting standard normal distribution tables, resulting in a probability of 0.4129.

Explanation:

The task at hand is to calculate the probability that the sample mean income is less than 42 thousand dollars. Given the population mean (μ) of 43 thousand dollars and the standard deviation (σ) of 29 thousand dollars, with a sample size (n) of 41, we can use the Central Limit Theorem to determine that the sampling distribution of the sample mean will be normally distributed with a mean equal to the population mean and a standard deviation equal to the population standard deviation divided by the square root of the sample size.

To find the probability, we first calculate the standard error (SE), which is:

SE = σ / √n = 29 / √41 ≈ 4.53

We then calculate the z-score for the sample mean of 42 thousand dollars as:

Z = (X - μ) / SE = (42 - 43) / 4.53 ≈ -0.22

Using standard normal distribution tables or a calculator, the probability of a z-score being less than -0.22 is approximately 0.4129. Thus, the answer is B) 0.4129.

Answer:

At least  one of the events A and B occurs:

(a) 0.5

(b) 0.44

All of the events A, B, C occur:

(a) 0.024

(b) 0

Step-by-step explanation:

Given:

P(A) = 0.2, P(B) = 0.3, P(C) = 0.4

At least  one of the events A and B occurs:

(a) If A and B are mutually exclusive events, then their intersection is 0.

Probability of at least one of them occurring means either of the two occurs or the union of the two events. Therefore,

[tex]P(A\ or\ B)=P(A)+P(B)-P(A\cap B)\\P(A\ or\ B)=0.2+0.3-0=0.5[/tex]

(b) If A and B are independent events, then their intersection is equal to the product of their individual probabilities. Therefore,

[tex]P(A\ or\ B)=P(A)+P(B)-P(A\cap B)\\P(A\ or\ B)=P(A)+P(B)+P(A)P(B)\\P(A\ or\ B)=0.2+0.3-(0.2)(0.3)=0.5-0.06=0.44[/tex]

All of the events A, B, C occur together:

(a) All of the events occurring together means the intersection of all the events.

If A, B, and C are independent events, then their intersection is equal to the product of their individual probabilities. Therefore,

[tex]P(A\cap B\cap C)=P(A)\cdot P(B)\cdot P(C)\\P(A\cap B\cap C)=0.2\times 0.3\times 0.4=0.024[/tex]

(b) If A, B, and C are mutually exclusive events means events A, B, and C can't happen at the same time. Therefore, their intersection is 0.

[tex]P(A\cap B\cap\ C)=0[/tex]

Probability that at least one of the events A and B occurs:

(a) When A and B are mutually exclusive: 0.5 (b) When A and B are independent: 0.44

Probability that all of the events A, B, and C occur:

(a) When A, B, and C are independent: 0.024 (b) When A, B, and C are mutually exclusive: 0

To solve these probability questions, let's understand the given information and the rules of probability related to mutually exclusive and independent events.

Let P(A) = 0.2, P(B) = 0.3, and P(C) = 0.4.

1. Finding the probability that at least one of the events A and B occurs:

(a) When A and B are mutually exclusive:

Mutually exclusive events mean that A and B cannot occur at the same time. Therefore, the probability that at least one of them occurs is given by:

P(A OR B) = P(A) + P(B)

So, P(A OR B) = 0.2 + 0.3 = 0.5

(b) When A and B are independent:

Independent events mean that the occurrence of one event does not affect the occurrence of the other. The probability that at least one of them occurs is given by:

P(A OR B) = P(A) + P(B) - P(A AND B)

For independent events, P(A AND B) is calculated as: P(A AND B) = P(A) × P(B)

So, P(A AND B) = 0.2 × 0.3 = 0.06

Therefore, P(A OR B) = 0.2 + 0.3 - 0.06 = 0.44

2. Finding the probability that all of the events A, B, and C occur:

(a) When A, B, and C are independent:

For independent events, the probability that all of them occur is given by:

P(A AND B AND C) = P(A) × P(B) × P(C)

So, P(A AND B AND C) = 0.2 × 0.3 × 0.4 = 0.024

(b) When A, B, and C are mutually exclusive:

If events are mutually exclusive, it means they cannot all occur together. Therefore, the probability that all of them occur is:

P(A AND B AND C) = 0

Answer:

A. TYPE 1 ERROR only

Step-by-step explanation:

In general terms:

‘a hypothesis has been rejected when it should have been accepted’. When this occurs, it is called a type I error, and,

‘a hypothesis has been accepted when it should have been rejected’.

When this occurs, it is called a type II error,

When testing a hypothesis, the largest value of probability which is acceptable for a type I error is called the level of significance of the test. The level of significance is indicated by the symbol α (alpha) and the levels commonly adopted are 0.1,0.05,0.01, 0.005 and 0.002.

A level of significance of 1%,say,0.01 means that 1 times in 100 the hypothesis has been rejected when it should have been accepted.

In significance tests, the following terminology is frequently adopted:

(i) if the level of significance is 0.01 or less, i.e. the confidence level is 9 9% or more, the results are considered to be highly significant, i.e. the results are considered likely to be correct,

(ii) if the level of significance is 0.05 or between 0.05and0.01,i.e.theconfidencelevelis95%or between 95% and 99%, the results are considered to be probably significant, i.e. the results are probably correct,

(iii) if the level of significance is greater than 0.05, i.e. the confidence level is less than 95%, the results are considered to be not significant, that is, there are doubts about the correctness of the results obtained.

Answer:

Chrissy walked 1/4 miles on Monday and Tuesday combined.

Step-by-step explanation:

1/8+1/8=2/8

2/8=1/4 you take half of both sides to get the lowest terms using simplification.

Answer:

45/100 = 0.45

Step-by-step explanation:

Divide 9 by 20.

9/20 = 0.45

Now write 36/100 as a decimal.

36/100 = 0.36

Write 45/100 as a decimal: 0.45

Answer: 45/100 = 0.45

64

Step-by-step explanation:

We will divide the volume of the larger cube with that if the smaller cube. However, we’ll first have to convert them to the same SI units;

12 inches  = 1 foot

Therefore the volume of the larger cube;

12 * 12 * 12 = 1728 inches cubed

The volume of the smaller cube;

3 * 3 * 3 = 27 inches cubed

Divide the two;

1728/27

= 64

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Answer:

E. None of the above answers is correct

Step-by-step explanation:

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"

If we assume that we have [tex]5[/tex] groups and on each group from [tex]j=1,\dots,20[/tex] we have [tex]20[/tex] individuals on each group we can define the following formulas of variation:  

[tex]SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2 [/tex]

[tex]SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 [/tex]

[tex]SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2 [/tex]

And we have this property

[tex]SST=SS_{between}+SS_{within}[/tex]

The degrees of freedom for the numerator on this case is given by [tex]df_{num}=df_{within}=k-1=5-1=4[/tex] where k =5 represent the number of groups.

The degrees of freedom for the denominator on this case is given by [tex]df_{den}=df_{between}=N-K=5*20-5=95[/tex].

And the total degrees of freedom would be [tex]df=N-1=5*20 -1 =99[/tex]

On this case the correct answer would be 4 for the numerator and 95 for the denominator.

E. None of the above answers is correct

Answer:

Option A) One tailed test is a hypothesis test in which rejection region is in one tail of the sampling distribution

Step-by-step explanation:

One Tailed Test:

[tex]H_0: \bar{x} = \mu\\H_A: \bar{x} < \mu\text{ or } \bar{x} > \mu[/tex]

Thus, for one tailed test,

Option A) One tailed test is a hypothesis test in which rejection region is in one tail of the sampling distribution

Final answer:

A one-tailed test is a hypothesis test where the rejection region is in one tail of the sampling distribution, and is used when the research hypothesis is directional, testing the possibility of a relationship in one direction only. (Option a)

Explanation:

A one-tailed test is a hypothesis test in which the rejection region is in one tail of the sampling distribution. This means that the correct answer to the student's question is a. hypothesis test in which rejection region is in one tail of the sampling distribution. In a one-tailed test, you are testing for the possibility of the relationship in one direction and completely disregarding the possibility of a relationship in the other direction (which would be the criterion of a two-tailed test).

Depending on the research hypothesis, the rejection region could be in the lower tail (option c) or the upper tail (option d) of the probability distribution. For example, if the alternative hypothesis is HA: X > μ, we are performing a right-tailed test and rejects the null hypothesis if the test statistic falls in the upper tail (right side) of the distribution. If the alternative hypothesis is HA: X < μ, it is a left-tailed test and the rejection region would be in the lower tail (left side).

Final answer:

The height of the tree is determined by the angles of elevation and the distance between points A and B on the ground. Given the 45° and 30° angles of elevation and knowing that the horizontal distance AB is 100 feet, we can infer that the tree height is also 100 feet.

Explanation:

To solve for the height of the tree, we can use the trigonometric properties of right triangles and the given angles of elevation. Since the angles of elevation from points A and B to the top of the tree T are 45° and 30° respectively, and the horizontal distance between A and B is 100 feet, we know that we have two separate right triangles with angle T being the top of the tree and O being the base of the tree. We also have an angle of 60° for ∠AOB which will help us determine the distances OA and OB.

Using the 45° angle from point A, we have a 45°-45°-90° triangle, which means the height of the tree (OT) is equal to OA, the distance from the base of the tree to point A. From point B with a 30° angle, we have a 30°-60°-90° triangle. In a 30°-60°-90° triangle, the height (OT) would be √3 times smaller than the distance OB (the longer side).

Since ∠AOB is 60° we know that triangle AOB is equilateral. Therefore, distances OA and OB are both 100 feet. Now we just need to calculate the height OT in one of the triangles. Taking the 45° triangle, we have OT = OA = 100 feet. Thus, the height of the tree T is 100 feet.

Final answer:

The height of the tree can be determined by using right triangle trigonometry involving the angles of elevation from two points on the ground and the known distance between them. Mathematically, the height of the tree is found to be approximately 28.87 feet.

Explanation:

To determine the height of the tree, we use the provided angles of elevation from points A and B and the distance between these points. The tree, points A and B, and the angles form two right triangles, one with a 45° angle of elevation and the other with a 30° angle of elevation. Since the angle of elevation from point A is 45°, we know that for a 45° right-angled triangle, the tangent is 1, which means that the opposite side (height of the tree) is equal to the adjacent side (distance from A to O).

First, we can calculate the distance from A to O using the angle of 60° between points A and O given as ∠AOB. The distance AO is half of AB because ∠AOB is an equilateral triangle's angle so AO = 50 feet. Next, using the fact that the tan 30° from point B is equal to the height of the tree divided by the distance from B to O, we can calculate the height. Since the distance from A to O is 50 feet, the distance from B to O is also 50 feet. The tangent of 30° is 1/√3, so the tree's height (H) is equal to 50 feet times tan 30°, which is approximately 28.87 feet.

The critical t values are reject the null hypothesis and shut down the process.

The correct option is (A).

To find the critical t values for a one-tailed t-test with a significance level of [tex]\( \alpha = 0.10 \)[/tex], we need to look up the t-distribution table or use statistical software.

Since we have a sample size of 17, the degrees of freedom ( df ) for the t-test will be [tex]\( n - 1 = 17 - 1 = 16 \).[/tex]

For a one-tailed t-test with a significance level of [tex]\( \alpha = 0.10 \)[/tex] and 16 degrees of freedom, we find the critical t value.

From the t-distribution table or using statistical software, the critical t value for [tex]\( \alpha = 0.10 \) and \( df = 16 \)[/tex] is approximately 1.337.

So, the critical t values are approximately [tex]\( \pm 1.337 \).[/tex]

Now, we compare the calculated t value for the sample mean to these critical values. If the calculated t value falls beyond these critical values, we reject the null hypothesis.

Since the problem does not provide the calculated t value, we cannot determine whether to reject the null hypothesis or not.

However, based on the information given, we can see that the mean weight of the USB flash drives in the last sample is 31.9 grams, which is heavier than the expected mean weight of 30 grams. If the calculated t value corresponds to a significantly larger value than the critical t value, we may reject the null hypothesis and conclude that the mean weight is significantly different from 30 grams, potentially leading to shutting down the process for adjustment.

Therefore, the correct answer would be:

A. reject the null hypothesis and shut down the process.

Since |4.35| > 1.746, we reject the null hypothesis and conclude that the mean weight of the USB flash drives is significantly different from 30 grams. Therefore, the correct answer is: A. reject the null hypothesis and shut down the process.

To find the critical t-values for a significance level of [tex]\( \alpha = 0.10 \)[/tex] and a sample size of n = 17, we can use a t-distribution table or a statistical software.

For a two-tailed test at a significance level of [tex]\( \alpha = 0.10 \)[/tex] and degrees of freedom df = n - 1 = 17 - 1 = 16, we need to find the critical t-values that correspond to the upper and lower tails of the t-distribution.

Using a t-distribution table or a statistical software, the critical t-values for a significance level of [tex]\( \alpha = 0.10 \)[/tex] and 16 degrees of freedom are approximately [tex]\( t_{\alpha/2} = \pm 1.746 \).[/tex]

So, the critical t-values are approximately[tex]\( \pm 1.746 \).[/tex]

Now, let's compare the calculated t-value with the critical t-values:

The calculated t-value is given by:

[tex]\[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \][/tex]

Where:

- [tex]\( \bar{x} \)[/tex] is the sample mean (31.9 grams)

- [tex]\( \mu \)[/tex] is the population mean (30 grams)

- s is the sample standard deviation (1.8 grams)

- n is the sample size (17)

Substitute the given values:

[tex]\[ t = \frac{31.9 - 30}{\frac{1.8}{\sqrt{17}}} \]\[ t \approx \frac{1.9}{\frac{1.8}{4.123}} \]\[ t \approx \frac{1.9}{0.437} \]\[ t \approx 4.35 \][/tex]

Since |4.35| > 1.746, we reject the null hypothesis and conclude that the mean weight of the USB flash drives is significantly different from 30 grams. Therefore, the correct answer is:

A. reject the null hypothesis and shut down the process.

Answer:

The 95% confidence interval would be given by (12.03;14.37)    

Step-by-step explanation:

1) 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=13.2[/tex] represent the sample mean  

[tex]\mu[/tex] population mean (variable of interest)

s=18.9 represent the sample standard deviation

n=997 represent the sample size  

2) Calculate the confidence interval

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=997-1=996[/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 tabel to find the critical value. The excel command would be: "=-T.INV(0.025,996)".And we see that [tex]t_{\alpha/2}=1.96[/tex] and this value is exactly the same for the normal standard distribution and makes sense since the sample size is large enough to approximate the t distribution with the normal standard distribution.

Now we have everything in order to replace into formula (1):

[tex]13.2-1.96\frac{18.9}{\sqrt{997}}=12.03[/tex]    

[tex]13.2+1.96\frac{18.9}{\sqrt{997}}=14.37[/tex]

So on this case the 95% confidence interval would be given by (12.03;14.37)    

Answer:

4 to 34

Step-by-step explanation:

Answer:

There are a few correct equivalents to this. Please mark brainliest!!!

Step-by-step explanation:

4:34, 6:51, 8:68, and 10;85

Answer:

y+19

Step-by-step explanation:

simply because when it says "more than" or "less than" you have to flip it so it wouldnt be 19+y.

Answer:

The 99% confidence interval for the mean would be (301.064;333.336) mA  

Step-by-step explanation:

1) Notation and some definitions

n=10 sample selected

[tex]\bar x=317.2mA[/tex] sample mean for the sample tubes selected

[tex]s=15.7mA[/tex] sample deviation for the sample selected

Confidence = 99% or 0.99

[tex]\alpha=1-0.99=0.01[/tex] significance level

A confidence interval for the mean is used to "places boundaries around an estimated [tex]\bar X[/tex] so that the true population mean [tex]\mu[/tex] would be expected to lie within those boundaries with a confidence specified. If the uncertainty is large, then the interval between the boundaries must be wide; if the uncertainty is small, then the interval can be narrow"

2) Formula to use

For this case the sample size is <30 and the population standard deviation [tex]\sigma[/tex] is not known, so for this case we can use the t distributon to calculate the critical value. The first step would be calculate alpha

[tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2=0.005[/tex], then we can calculate the degrees of freedom given by:

[tex]df=n-1=10-1=9[/tex]

Now we can calculate the critical value [tex]t_{\alpha/2}=3.25[/tex]

And then we can calculate the confidence interval with the following formula

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]  (1)

3) Calculate the interval

Using the formula (1) and replacing the values that we got we have:

[tex]317.2 - 3.25\frac{15.7}{\sqrt{10}}=301.064[/tex]  

[tex]317.2 + 3.25\frac{15.7}{\sqrt{10}}=333.336[/tex]

So then the 99% confidence interval for the mean would be (301.064;333.336)mA  

Interpretation: A point estimate for the true mean of brightness level for the tubes in the population is 317.2mA, and we are 99% confident that the true mean is between 301.064 mA and 333.336 mA.

Answer:

0.5746

Step-by-step explanation:

Lets call X monthly bill chosen. X is a random variable with distribution N(μ=53,σ =15). We first standarize X, lets call W = (X-53)/15, random variable obtained from X by substracting μ and dividing by σ. W is a random variable with distribution N(0,1) and the cummulative function of W, Φ, is tabulated. You can find the values of Φ on the attached file. Using this information we obtain

[tex] P(47 < X < 74) = P(\frac{47-53}{15} < \frac{X-53}{15} < \frac{74-53}{15}) = P(-0.4 < W < 1.4) = \phi(1.4) - \phi(-0.4) [/tex]

A standard normal random variable has a symmetric density function, as a result Φ(-0.4) = 1- Φ(0.4) = 1- 0.6554 = 0.3446. Also, Φ(1.4) = 0.9192. We conclude that Φ(1.4)-Φ(-0.4) =  0.9192-0.3446 = 0.5746, that is the probability that the phone bill is between 47 and 74 euros.

Answer:

a. categorical

b. 1018

c. 28.98%

d.  Most voters interviewed are against the new tax

Step-by-step explanation:

Hello!

There was a poll made to know the voter opinion on a law that would increase the gas tax to 20 cents per gallon, this money is going to use for road and bridges improvement and build more mass transportation in the state.

Data:

295 support

672 against

51 no opinion

a. The study variable is "Opinion of the votes on the new gas tax" Categorized: "support; against; no opinion"

This is a categorical variable, that can take one number on a limited amount of possibles values, assigning each observation to a group of nominal categories based on a qualitative feature.

In this case, each voter is assigned to a category according to their opinion on the new tax.

b. To know the sample size you have to add the subtotals of each category:

n= 295 + 672 + 51 = 1018

c. To calculate the percentage of people supporting the new tax, you have to divide the subtotal by the total and multiply it by 100

[tex](\frac{295}{1018})*100[/tex] = 28.978% ≅ 28.98%

d. To answer this it's better to calculate the percentage of each category:

Support: 28.98%

Against: 66.01%

No opinion: 5.01%

Since the category "against" has the greater percentage, you can say that most voters interviewed are against the new gas tax

I hope it helps!

Answer:

0.191 is the probability that a group of 12 randomly selected applicants would have a mean SAT score that is greater than 525 but below the current admission standard of 584.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 500

Standard Deviation, σ = 100

n = 12

We are given that the distribution of SAT score is a bell shaped distribution that is a normal distribution.

Formula:

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

P(greater than 525 but 584)

Standard error due to sampling =

[tex]\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{100}{\sqrt{12}}[/tex]

[tex]P(525 < x < 584) = P(\displaystyle\frac{525 - 500}{\frac{100}{\sqrt{12}}} < z < \displaystyle\frac{584-500}{\frac{100}{\sqrt{12}}}) = P(0.866 < z < 2.909)\\\\= P(z \leq 2.909) - P(z < 0.866)\\= 0.998 - 0.807 = 0.191 = 19.1\%[/tex]

[tex]P(525 < x < 584) = 19.1\%[/tex]

0.191 is the probability that a group of 12 randomly selected applicants would have a mean SAT score that is greater than 525 but below the current admission standard of 584.

Country: Cultural issues, currency risks. Region: Land costs, air and rail systems. (Total: 10 words)

In assessing country and region success factors, certain variables hold significant weight. Country success factors encompass aspects such as cultural issues and currency risks, which directly influence a nation's economic stability and investment attractiveness.

Cultural nuances affect business practices and consumer behaviors, while currency risks impact trade competitiveness and profitability.

Conversely, region success factors pivot around considerations like land costs and transportation infrastructure, particularly air and rail systems. Land costs dictate the feasibility of establishing operations or acquiring property within a region, shaping investment decisions.

Meanwhile, efficient air and rail systems facilitate logistical operations and market access, enhancing the competitiveness of businesses located within the region.

These factors collectively contribute to a region's appeal for businesses seeking to establish or expand operations.

Thus, understanding and appropriately navigating these variables are crucial for organizations aiming to optimize their strategic positioning at both the national and regional levels, ensuring long-term success and sustainability in an increasingly complex global landscape.

Answer:

We use the Binomial Distribution   where p =.84 then probability will be 0.931.

Step-by-step explanation:

P(X≥ 2) = P(X=2) + P (X=3)

           = (³₂)(.84)^2(.16) +  (³3)(.84)^3(.16)⁰

           ≅ 0.931

Hence 0.931 is the answer.

The probability that at least two out of three items chosen randomly are in perfect condition, given that the chance for a randomly chosen item being perfect is 84%, is 97% or 0.97 in decimal form.

This is a question about the probability that we call binomial probability. Here we have to find the probability that at least two out of three randomly chosen items are in perfect condition while the likelihood of a random item being in perfect condition is 84%.

There are two scenarios in which the lot will be accepted: (1) exactly two of the products are perfect, and (2) all three products are perfect.

1. Two products are perfect (using binomial probability formula):

Prob(Two perfect) = 3C2 * (0.84)^2 * (1-0.84)^1 = 0.38

2. All three are perfect:

Prob(Three perfect) = 3C3 * (0.84)^3 * (1 - 0.84)^0 = 0.59

Thus, adding these probabilities gives the total probability that either two or three products are in good condition, which would result in the lot being accepted:

Total Probability = 0.38 + 0.59 = 0.97 or 97%

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Answer: (47.51, 54.49)

Step-by-step explanation:

Confidence interval for population mean is given by :-

[tex]\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]

, where n= sample size .

[tex]\sigma[/tex] = population standard deviation.

[tex]\overline{x}[/tex] = sample mean

[tex]z_{\alpha/2}[/tex] = Two -tailed z-value for [tex]{\alpha[/tex] (significance level)

As per given , we have

[tex]\sigma=11.8\text{ ounces}[/tex]

[tex]\overline{x}=51 \text{ ounces}[/tex]

n= 44

Significance level for 95% confidence = [tex]\alpha=1-0.95=0.05[/tex]

Using z-value table ,

Two-tailed Critical z-value : [tex]z_{\alpha/2}=z_{0.025}=1.96[/tex]

Now, the 95% confidence interval for the true population mean textbook weight will be :-

[tex]51\pm (1.96)\dfrac{11.8}{\sqrt{44}}\\\\=51\pm(1.96)(1.7789)\\\\=51\pm3.486644\approx51\pm3.49\\\\=(51-3.49,\ 51+3.49)\\\\=(47.51,\ 54.49) [/tex]

Hence, the 95% confidence interval for the true population mean textbook weight. :  (47.51, 54.49)

Answer: 99% confidence interval would be [tex](1346.37,1553.63)[/tex]

Step-by-step explanation:

Since we have given that

Sample size = 30

Sample mean = $1450

Standard deviation = $220

We need to find the 99% confidence interval.

So, z = 2.58

So, confidence interval would be

[tex]\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=1450\pm 2.58\times \dfrac{220}{\sqrt{30}}\\\\=1450\pm 103.63\\\\=(1450-103.63,1450+103.63)\\\\=(1346.37,1553.63)[/tex]

Hence, 99% confidence interval would be [tex](1346.37,1553.63)[/tex]

(a) False. The sum will not be around 1, give or take 1.   (b) False. The probability that the sum will be in the range 0 to 2 is not about 68%.

(a).

Expected sum = (99/100 x 0) + (1/100 x  1)

= 1/100

= 0.01

Therefore, the expected value of the sum is 0.01, which is much closer to 0 than 1. While it is possible to observe a sum of 1 or close to 1 in a particular set of draws, it is not likely to be the average or typical outcome.

(b) False. The probability that the sum will be in the range 0 to 2 is not about 68%.

To determine the probability, to consider the possible outcomes that result in a sum between 0 and 2.

The possible outcomes are:

Therefore, the total probability of the sum being in the range 0 to 2 is approximately:

[tex]0.366 + 0.369 + 0.185 = 0.92,[/tex]

which is far greater than 68%.

Learn more about probability here:

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