Consider an airfoil in a wind tunnel (i.e., a wing that spans the entire test section). Prove that the lift per unit span can be obtained from the pressure distributions on the top and bottom walls of the wind tunnel (i.e., from the pressure distributions on the walls above and below the airfoil).

Step-by-step explanation:

If x is her score on the final exam, then the average is:

(74 + 88 + 61 + 83 + 2x) / 6

(306 + 2x) / 6

51 + ⅓x

We want this to be greater than or equal to 76.

51 + ⅓x ≥ 76

⅓x ≥ 25

x ≥ 75

In order to get an average of 76 or higher, Shureka's score on the final exam must be greater than or equal to 75.

Answer:

la familia invierte 8.33% de los ingresos totales en ocio

Step-by-step explanation:

Representando los ingresos totales por I:

- Gastos de funcionamiento = 2/3*I

- Ahorro  = 1/4*I

- En ocio : lo que resta = I - 2/3*I - 1/4*I = I - 11/12*I = 1/12*I (8.33% de I)

por lo tanto la familia invierte 8.33% de los ingresos totales en ocio

Answer:

The probability of selecting 5 female and 2 male students is 0.052.

Step-by-step explanation:

The class comprises of 7 female students and 10 male students.

Total number of students: 17.

Number of female students, 7.

Number of male students, 10.

The probability of an event E is:

[tex]P(E)=\frac{Favorable\ outcomes}{Total\ number\ of] outcomes}[/tex]

The number of ways to select 7 students from 17 is:

[tex]N ={17\choose 7}=\frac{17!}{7!(17-7)!}= 19448[/tex]

The number of ways to select 5 female students of 7 females is:

[tex]n(F) ={7\choose 5}=\frac{7!}{5!(7-5)!}= 21[/tex]

The number of ways to select 2 male students of 10 males is:

[tex]n(M) ={10\choose 2}=\frac{10!}{2!(10-2)!}= 45[/tex]

Compute the probability of selecting 5 female and 2 male students as follows:

P (5 F and 2 M) = [n (F) × n (M)] ÷ N

                         [tex]=\frac{21\times45}{19448} \\=0.05183\\\approx0.052[/tex]

Thus, the probability of selecting 5 female and 2 male students is 0.052.

Answer:

a) left handed people

b) Binomial probability distribution with pdf

[tex]P(X=x)=15Cx0.1^{x} 0.9^{15-x}[/tex]

where x=0,1,2,...,15.

c) Histogram is attached

d) The shape of histogram depicts that distribution is rightly skewed.

e) 1.5

f) 1.35

g) 1.16

Step-by-step explanation:

a)

The random variable in the given scenario is " left handed people"

b)

The scenario represents the binomial probability distribution as the outcome is divided into one of two categories and experiment is repeated fixed number of times i.e. 15 and trails are independent. The pdf of binomial distribution is

[tex]P(X=x)=nCxp^{x} q^{n-x}[/tex]

Here n=15, p=0.1 and q=1-p=0.9.

So, the pdf would be

[tex]P(X=x)=15Cx0.1^{x} 0.9^{n-x}[/tex]

where x=0,1,2,...,15.

c)

Histogram is constructed by first computing probabilities on all x points i.e. x=0, x=1 , .... ,x=15 and then plotting all probabilities with respective x values. Histogram is in attached image.

d)

The tail of histogram is to the right side and thus the histogram depicts that given probability distribution is rightly skewed.

e)

The mean of binomial probability distribution is computed by multiplying number of trails and probability of success.

mean=np=15*0.1=1.5

f)

The variance of binomial probability distribution is computed by multiplying number of trails and probability of success and probability of failure.

variance=npq=15*0.1*0.9=1.35

g)

The standard deviation can be calculated by simply taking square root of variance

S.D=√npq=√1.35=1.16

The proportion of left-handed people follows a binomial distribution

The given parameters are:

[tex]\mathbf{n = 15}[/tex] -- the sample size

[tex]\mathbf{p = 10\%}[/tex] --- the proportion of left-handed people

(a) The random variable

The distribution is about left-handed people.

Hence, the random variable is left-handed people

(b) The probability distribution

If the proportion of left-handed people is 10%, then the proportion of right-handed people is 90%.

So, the probability distribution function is:

[tex]\mathbf{P(x) = ^nC_x p^x (1 - p)^{n -x}}[/tex]

This gives

[tex]\mathbf{P(x) = ^nC_x (10\%)^x (1 - 10\%)^{n -x}}[/tex]

[tex]\mathbf{P(x) = ^nC_x 0.1^x 0.9^{n -x}}[/tex]

Hence, the probability distribution function is [tex]\mathbf{P(x) = ^nC_x 0.1^x 0.9^{n -x}}[/tex]

(c) The histogram

To do this, we calculate P(x) for x = 0 to 15

[tex]\mathbf{P(0) = ^{15}C_0 \times 0.1^0 \times 0.9^{15 -0} = 0.206}[/tex]

[tex]\mathbf{P(1) = ^{15}C_1 \times 0.1^1 \times 0.9^{15 -1} = 0.343}[/tex]

.....

..

[tex]\mathbf{P(15) = ^{15}C_{15} \times 0.1^{15} \times 0.9^{15 -15} = 10^{-15}}[/tex]

See attachment for the histogram

(d) The mean

This is calculated as:

[tex]\mathbf{\bar x = np}[/tex]

So, we have:

[tex]\mathbf{\bar x = 15 \times 10\% }[/tex]

[tex]\mathbf{\bar x= 1.5}[/tex]

Hence, the mean is 1.5

(e) The variance

This is calculated as:

[tex]\mathbf{Var = np(1 - p)}[/tex]

So, we have:

[tex]\mathbf{Var = 15 \times 10\% \times (1 - 10\%)}[/tex]

[tex]\mathbf{Var = 1.35}[/tex]

Hence, the variance is 1.35

(f) The standard deviation

This is calculated as:

[tex]\mathbf{\sigma = \sqrt{Var}}[/tex]

So, we have:

[tex]\mathbf{\sigma = \sqrt{1.35}}[/tex]

[tex]\mathbf{\sigma =1.16}[/tex]

Hence, the standard deviation is 1.16

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Answer:

This process is out of control.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

A probability is said to be unusual if it's z-score has a pvalue of 0.05 or lower, or a pvalue of 0.95 or higher.

In this problem, we have that:

[tex]\mu = 10.2 \sigma = 1.04[/tex]

If a randomly selected minute has 13.9 hits, would the process be considered in control or out of control?

The process will be considered out of control if it's z-score(Z when X = 13.9) has a pvalue of 0.95 or higher. Otherwise(it will be positive, since 13.9 is above the mean), it will be considered in control.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{13.9 - 10.2}{1.04}[/tex]

[tex]Z = 3.56[/tex]

[tex]Z = 3.56[/tex] has a pvalue of 0.9999. So there is only a 1-0.9999 = 0.0001 = 0.01% probability of getting 13.9 hits a minute.

So this process is out of control.

If using the empirical rule, since 13.9 hits is more than two standard deviations above the mean of 10.2 hits, the process could be considered out of control. However, establishing specific control limits is necessary for a definitive answer.

To determine whether a process is in control or out of control, we assess whether a given measurement is within the expected range of a normal distribution, often using the empirical rule or control limits. Given that the process has a mean of 10.2 hits per minute and a standard deviation of 1.04 hits, under the empirical rule, approximately 95% of the data should fall within two standard deviations of the mean (that is, between roughly 8.12 and 12.28 hits).

With 13.9 hits in a randomly selected minute, this count is significantly more than two standard deviations above the mean, suggesting that the process might be out of control. However, to make a definitive statement about control status, specific control limits must be established, often based on the particular specifications of the process being monitored.

Answer:

If 40 percent of the English-speaking men at the hotel are English and 60 percent are American, the the probability that the writer is an Englishman

 is 40% or 0.4.

Step-by-step explanation:

i) If 40 percent of the English-speaking men at the hotel are English and 60 percent are American, the the probability that the writer is an Englishman

 is 40% or 0.4.

The probability that the writer is an Englishman given that a vowel was chosen from his spelling is approximately 45.45%. This was calculated using the respective probabilities of selecting a vowel from the words 'rigour' and 'rigor' and Bayes' Theorem.

Given the scenario where a man writes a word, and a letter taken at random is a vowel, we aim to determine the probability that the writer is an Englishman.

The words are:

Letters 'i', 'o', and 'u' are the vowels. Let's calculate the likelihood of selecting a vowel from each spelling:

Using Bayes' Theorem, let's find the probability the writer is English (E), given a vowel (V) was selected:

P(E|V) = [P(V|E) * P(E)] / [P(V|E) * P(E) + P(V|A) * P(A)]

Where:

So:

P(E|V) = [0.5 * 0.4] / [0.5 * 0.4 + 0.4 * 0.6] = 0.2 / (0.2 + 0.24) = 0.2 / 0.44 ≈ 0.4545

Therefore, the probability that the writer is an Englishman is approximately 0.4545 or 45.45%.

Answer:

The possible number of ways to select distinct (a, b) such that (a + b) is even is 534.

Step-by-step explanation:

The range 1 - 99 has 99 numbers, since 1 and 99 are inclusive.

Of these 50 numbers are odd and 49 are even.

The two distinct numbers a and b must have an even sum and a should be a multiple of 9.

The sum of two numbers is even only when both are odd or both are even.

The possible values that a can assume are,

a = {9, 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99}

Thus, a can assume 6 odd values and 5 even values.

Total number of ways to select a and b such that both are odd and their sum is even is:

[tex]n(Odd\ a\ and\ b)=n(Odd\ value\ of\ a)\times n(Odd\ value\ of\ b)=6\times49=294[/tex]

Total number of ways to select a and b such that both are even and their sum is even is

[tex]n(Even\ a\ and\ b)=n(E\ value\ of\ a)\times n(Even\ value\ of\ b)=5\times48=240[/tex]

Total number of ways to select distinct (a, b) such that (a + b) is even is =

[tex]=n(Odd\ a\ and\ b)+n(Even\ a\ and\ b)=294+240=534[/tex]

Thus, the possible number of ways to select distinct (a, b) such that (a + b) is even is 534.

Answer:

[tex]0.753 - 1.64\sqrt{\frac{0.753(1-0.753)}{1000}}=0.731[/tex]

[tex]0.753 + 1.64\sqrt{\frac{0.753(1-0.753)}{1000}}=0.775[/tex]

The 95% confidence interval would be given by (0.731;0.775)

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".  

The population proportion have the following distribution

[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by [tex]\alpha=1-0.90=0.1[/tex] and [tex]\alpha/2 =0.005[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64[/tex]

The confidence interval for the mean is given by the following formula:  

[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]

The estimated proportion for this case is:

[tex] \hat p = \frac{X}{n}= \frac{753}{1000}=0.753[/tex]

If we replace the values obtained we got:

[tex]0.753 - 1.64\sqrt{\frac{0.753(1-0.753)}{1000}}=0.731[/tex]

[tex]0.753 + 1.64\sqrt{\frac{0.753(1-0.753)}{1000}}=0.775[/tex]

The 95% confidence interval would be given by (0.731;0.775)

Answer:

There is a 90.46% probability that it was repaired by a competent shop, given that it was repaired correctly.

Step-by-step explanation:

We have these following probabilities:

An 87% probability that an air conditioner repair shop is competent.

A 13% probability that an air conditioner repair shop is incompetent.

An 85% probability that an compotent shop can repair the air.

A 60% probability than an incompetent shop can repair the air.

This can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened?

It can be calculated by the following formula

[tex]P = \frac{P(B).P(A/B)}{P(A)}[/tex]

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

In this problem we have that:

Probability that it was repaired by a competent shop, given that it was repaired correctly.

P(B) is the probability that it was repaired by a competent shop. 87% of the shops are competent, so [tex]P(B) = 0.87[/tex]

P(A/B) is the probability that it was repaired correctly, given that it was repaired by a competent shop. There is an 85% probability that an compotent shop can repair the air. So [tex]P(A/B) = 0.85[/tex]

P(A) is the probability that an air was repaired correctly.

This is 85% of 87% and 60% of 13%. So

[tex]P(A) = 0.85*0.87 + 0.60*0.13 = 0.8175[/tex]

Finally

[tex]P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.87*0.85}{0.8175} = 0.9046[/tex]

There is a 90.46% probability that it was repaired by a competent shop, given that it was repaired correctly.

Quota sampling is most commonly used in surveys. It is a non-probability sampling technique where researchers select individuals who meet certain criteria to be included in the sample.

Quota sampling is most commonly used in surveys. It is a non-probability sampling technique where researchers select individuals who meet certain criteria to be included in the sample.

In quota sampling, the researcher sets quotas or targets for each subgroup they want to include in the sample based on certain characteristics. For example, if the researcher wants equal representation of males and females in the sample, they would set quotas for each gender and continue selecting individuals until the quotas are met.

Quota sampling is often used when it is not possible or practical to obtain a random sample, but the researcher still wants to ensure representation of different subgroups within the sample.

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Answer:

ccvnnnrxfh koo HD fyu

Step-by-step explanation:

yyyhjkuz do Lakewooddrink instill k is s

Answer:

340 of the adults in the sample voted.

The 95% confidence interval estimate of the population percent of Americans that vote is (0.6391, 0.7209). This means that we are 95% sure that the true proportion of Americans that vote is between 0.6391 and 0.7209.

Step-by-step explanation:

In 2008, a random sample of 500 american adults was take and we found that 68% of them voted. How many of the 500 adults in the sample voted?

This is 68% of 500.

So 0.68*500 = 340.

340 of the adults in the sample voted.

Now construct a 95% confidence interval estimate of the population percent of Americans that vote and write a sentence to explain the confidence interval.

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence interval [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

Z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

For this problem, we have that:

[tex]n = 500, p = 0.68[/tex]

95% confidence interval

So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.68 - 1.96\sqrt{\frac{0.68*0.32}{500}} = 0.6391[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.68 + 1.96\sqrt{\frac{0.68*0.32}{500}} = 0.7209[/tex]

The 95% confidence interval estimate of the population percent of Americans that vote is (0.6391, 0.7209). This means that we are 95% sure that the true proportion of Americans that vote is between 0.6391 and 0.7209.

Answer:

Frist case: P=12/35

Second case: P=31/35

Step-by-step explanation:

An urn has 3 blue balls 4 red balls. 3 balls are drawn without replacement.

Frist case:

We calculate the number of possible combinations

{7}_C_{3}=\frac{7!}{3! · (7-3)!}=35

We calculate  the number of favorable combinations  

{3}_C_{2} · {4}_C_{1} =

=\frac{3!}{2! · (3-2)!} · \frac{4!}{1! · (4-1)!}

=3 · 4 = 12

Therefore, the probability is

P=12/35

Second case:

When we count on at least one ball to be blue, we go over the probability complement.

We calculate the probability that all the balls are red, then subtract this from 1.

We calculate the number of possible combinations

{7}_C_{3}=\frac{7!}{3! · (7-3)!}=35

We calculate  the number of favorable combinations  

{4}_C_{3}  = \frac{4!}{3! · (4-3)!=4

The probability is

P=4/35.

Therefore the probability on at least one ball to be blue

P=1-4/35

P=31/35

Answer: There is a 90% chance that the error range will be with in 3.29 ft² .

Step-by-step explanation:

Given : The  area of a parking lot is calculated to be 5,474 ft² .

Estimated standard deviation = 2 ft²

The critical z-value for 90% confidence interval is 1.645 (from z-table)

Then, the Maximum Anticipated Error = ( critical z-value ) x ( standard deviation )

= 1.645 (2) =3.29 ft²

i.e.  Maximum Anticipated Error = 3.29 ft²

Hence, there is a 90% chance that the error range will be with in 3.29 ft² .

Answer:

a) P ( A & B ) = 0.1995

b) P (A U B ) = 0.7905

c) P (A/B) = 0.2625

d) P(B/A')  = 0.194805

e) NOT mutually exclusive

f) NOT Independent

Step-by-step explanation:

Declare Events:

- buying gasoline = Event A

- buying groceries = Event B

Given:

- P(A) = 0.23

- P(B) = 0.76

- P(B/A) = 0.85

Find:

- a. Find the probability that a typical customer buys both gasoline and groceries.

- b. Find the probability that a typical customer buys gasoline or groceries.

- c. Find the conditional probability of buying gasoline given that the customer buys groceries.

- d. Find the conditional probability of buying groceries given that the customer did not buy gasoline.

- e Are these two events (groceries, gasoline) mutually exclusive?

- f  Are these two events independent?

Solution:

- a) P ( A & B ) ?

                     P ( A & B ) = P(B/A) * P(A) = 0.85*0.23 = 0.1995

- b) P (A U B ) ?

                    P (A U B ) = P(A) + P(B) - P(A&B)

                    P (A U B ) = 0.23 + 0.76 - 0.1995

                    P (A U B ) = 0.7905

- c) P ( A / B )?

                    P ( A / B ) = P(A&B) / P(B)

                                    = 0.1995 / 0.76

                                    = 0.2625

- d) P( B / A') ?

                   P( B / A') = P ( B & A') / P(A')

                   P ( B & A' ) = 1 - P( A / B) = 1 - 0.85 = 0.15

                   P ( B / A' ) = 0.15 / (1 - 0.23)

                                    = 0.194805

- e) Are the mutually exclusive ?

        The condition for mutually exclusive events is as follows:

                    P ( A & B ) = 0 for mutually exclusive events.

        In our case P ( A & B ) = 0.1995 is not zero.

        Hence, NOT MUTUALLY EXCLUSIVE

- f) Are the two events independent?

         The condition for independent events is as follows:

                    P ( A & B ) = P (A) * P(B) for mutually exclusive events.

        In our case,

                        0.1995 = 0.23*0.76

                        0.1995 = 0.1748 (NOT EQUAL)

        Hence, NOT INDEPENDENT

a. The probability a customer buys both gasoline and groceries is 0.23 * 0.85 = 0.1955.

b. The probability a customer buys either gasoline or groceries is 0.23 + 0.76 - 0.1955 = 0.7945.

c. Conditional probability of buying gasoline given groceries: 0.1955 / 0.76 ≈ 0.2572.

d. Conditional probability of buying groceries given no gasoline: 0.76 - 0.1955 / 0.77 ≈ 0.7331.

e. No, these events are not mutually exclusive.

f. No, these events are not independent.

Let's calculate the probabilities and answer each part of the question:

a. To find the probability that a typical customer buys both gasoline and groceries, you can use the formula for conditional probability:

P(Gasoline and Groceries) = P(Groceries | Gasoline) * P(Gasoline)

P(Gasoline and Groceries) = 0.85 * 0.23 = 0.1955

So, the probability that a typical customer buys both gasoline and groceries is 0.1955.

b. To find the probability that a typical customer buys gasoline or groceries, you can use the addition rule for probabilities:

P(Gasoline or Groceries) = P(Gasoline) + P(Groceries) - P(Gasoline and Groceries)

P(Gasoline or Groceries) = 0.23 + 0.76 - 0.1955 = 0.7945

So, the probability that a typical customer buys gasoline or groceries is 0.79

c. To find the conditional probability of buying gasoline given that the customer buys groceries, you can use the formula for conditional probability:

P(Gasoline | Groceries) = P(Gasoline and Groceries) / P(Groceries)

P(Gasoline | Groceries) = 0.1955 / 0.76 ≈ 0.2572

So, the conditional probability of buying gasoline given that the customer buys groceries is approximately 0.2572.

d. To find the conditional probability of buying groceries given that the customer did not buy gasoline, you can use the formula for conditional probability:

P(Groceries | No Gasoline) = P(Groceries and No Gasoline) / P(No Gasoline)

First, calculate P(No Gasoline):

P(No Gasoline) = 1 - P(Gasoline) = 1 - 0.23 = 0.77

Now, calculate P(Groceries and No Gasoline):

P(Groceries and No Gasoline) = P(Groceries) - P(Gasoline and Groceries) = 0.76 - 0.1955 = 0.5645

Now, find P(Groceries | No Gasoline):

P(Groceries | No Gasoline) = 0.5645 / 0.77 ≈ 0.7331

So, the conditional probability of buying groceries given that the customer did not buy gasoline is approximately 0.7331.

e. These two events (buying groceries and buying gasoline) are not mutually exclusive because it's possible for a customer to buy both groceries and gasoline, as we calculated in part (a).

f. To determine whether these two events are independent, we need to check if the conditional probabilities match the unconditional probabilities:

P(Gasoline | Groceries) = P(Gasoline) and P(Groceries | No Gasoline) = P(Groceries)

Let's check:

P(Gasoline | Groceries) ≈ 0.2572

P(Gasoline) = 0.23

P(Groceries | No Gasoline) ≈ 0.7331

P(Groceries) = 0.76

These conditional probabilities are not equal to the unconditional probabilities, so the events are not independent. In an independent event scenario, the conditional probabilities would be equal to the unconditional probabilities.

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a. To find the margin of error, we first calculate the standard error of the mean using the formula [tex]\(SE = \frac{s}{\sqrt{n}}\)[/tex], where s is the sample standard deviation and n is the sample size. Then, we use the formula for the margin of error [tex]\(ME = Z \times SE\)[/tex], where Z is the z-score corresponding to the desired level of confidence.

b. Once we have the margin of error, we can construct the confidence interval estimate of the population mean by adding and subtracting the margin of error from the sample mean.

c. To compare the prices for meals for two in mid-range restaurants in Hong Kong to those in Tokyo, we can use the confidence interval estimate of the population mean. If the confidence interval includes the mean price for Tokyo ($40), it suggests that there may not be a significant difference in prices between the two cities. However, if the confidence interval does not include $40, it suggests that there may be a significant difference in prices between the two cities.

Explanation:

a. Margin of error:

1. Calculate the sample mean [tex](\( \bar{x} \))[/tex] and sample standard deviation s from the given data.

2. Determine the sample size n.

3. Find the standard error of the mean SE using the formula [tex]\( SE = \frac{s}{\sqrt{n}} \)[/tex].

4. Look up the z-score corresponding to the desired level of confidence (e.g., 95%) from the standard normal distribution table.

5. Multiply the z-score by the standard error to find the margin of error [tex](\( ME \))[/tex].

b. Confidence interval estimate:

1. Calculate the margin of error ME.

2. Subtract the margin of error from the sample mean to find the lower bound of the confidence interval.

3. Add the margin of error to the sample mean to find the upper bound of the confidence interval.

c. Price comparison:

1. Check if the confidence interval estimate of the population mean includes the mean price for Tokyo ($40).

2. If the confidence interval includes $40, it suggests that there may not be a significant difference in prices between the two cities.

3. If the confidence interval does not include $40, it suggests that there may be a significant difference in prices between the two cities.

Answer: False

Step-by-step explanation:

In Inferential statistics the sample data is analysed rather than analysing the whole population, analysing the whole population may sometimes be impossible, like analysing the population of a whole country. Therefore, for inferential statistics, conclusions about the whole population are drawn by analyzing the corresponding sample data. The sample data are selected from the population and then analysed, the results can then be used to conclude on the whole population.