02.05) simplify i22. (1 point) 1 −1 −i i

Production of a gas flow meter takes place in two distinct operations. Measurements (n = 15) of the time required for the first

Karolina [17]

Answer:

a) P(a > 80) = 0.323

b) The 95% confidence interval = (73.40, 80.60)

c) The 95% confidence interval expresses that the mean of the distribution can always be found in the given range, with a 95% confidence level.

Step-by-step explanation:

X ~ (45, 4)

Y ~ (32, 2.5)

(X+Y) ~ (77, 6.5)

Let a = (X+Y)

a) Probability that the time required to complete both of those steps will exceed 80 min = P(a > 80)

This is a normal distribution problem

We then standardize 80 min time

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (a - μ)/σ = (80 - 77)/6.5 = 0.46

To determine the probability the time required to complete both of those steps will exceed 80 min = P(a > 80) = P(z > 0.46)

We'll use data from the normal probability table for these probabilities

P(a > 80) = P(z > 0.46) = 1 - P(z ≤ 0.46) = 1 - 0.677 = 0.323

b) 95% confidence interval for the expected total time required to produce one flow meter.

We need to obtain the margin of error

Margin of error = (critical value) × (standard error of the sample)

Critical value for a 95% confidence interval = critical value for a significance level of 5% = t(15-1, 0.05/2) = 2.145 (using the t-score since information on the population mean and standard deviation isn't known)

Standard error for the sample of sum of times = (standard deviation of the sum of times)/√n = (6.5/√15) = 1.678

Margin of error = 2.145 × 1.678 = 3.60

Limits of the confidence interval = (Sample mean ± margin of error)

Lower limit of the confidence interval = (Sample mean - margin of error) = 77 - 3.60 = 73.40

Upper limit of the confidence interval = 77 + 3.60 = 80.60

The confidence interval = (73.40, 80.60)

4 0

3 years ago

Suppose that a company has recently started manufacturing a new line of helicopters, and it took 106.496 labor days to complete

Allisa [31]

A "learning rate" expresses the ratio of the labor spent on the n-th item to that spent on the (n/2)-th item. That is, the 8th helicopter manufactured took days related by ...

0.8 × (days for 8th) = (days for 16th)

(days for 8th) = (days for 16th)/0.8 = 1.25×(days for 16th)

Then the days for the 4th are 1.25 times that, and the days for the 2nd are 1.25 times the days for the 4th. The days for the first helicopter will be

... 1.25⁴×106.496 days = 260 days

The appropriate choice is

... d. 260

5 0

2 years ago

A genetic experiment involving peas yielded one sample of offspring consisting of 437437 green peas and 129129 yellow peas. Use

navik [9.2K]

Answer:

The null hypothesis: $\mathbf{H_o: p=0.27}$

The alternative hypothesis: $\mathbf{H_1: p \neq 0.27}$

Test statistics : z = −2.30

P-value: = 0.02144

Decision Rule: Since the p-value is lesser than the level of significance; then we reject the null hypothesis.

Conclusion: We accept the alternative hypothesis and conclude that under the same circumstances the proportion of offspring peas will be yellow is not equal to 0.27

Step-by-step explanation:

From the given information:

Let's state the null and the alternative hypothesis;

Since The claim is that 27% of the offspring peas will be yellow.

The null hypothesis state that the proportion of offspring peas will be yellow is equal to 0.27.

i.e

$\mathbf{H_o: p=0.27}$

The alternative hypothesis state that the proportion of offspring peas will be yellow is not equal to 0.27

$\mathbf{H_1: p \neq 0.27}$

The test statistics:

we are given 437 green peas and 129 yellow apples;

Hence;

$\hat p = \dfrac{x}{n}$

where ;

$\hat p$ = sample proportion

x = number of success

n = total number of the sample size

$\hat p = \dfrac{129}{437+129}$

$\hat p = \dfrac{129}{566}$

$\mathbf{\hat p = 0.2279}$

Now; the test statistics can be computed as :

$z = \dfrac { \hat p -p }{\sqrt {\dfrac{p(1-p)}{n} } }$

$z = \dfrac {0.2279 -0.27 }{\sqrt {\dfrac{0.27(1-0.27)}{566} } }$

$z = \dfrac {-0.043 }{\sqrt {\dfrac{0.27(0.73)}{566} } }$

$z = \dfrac {-0.043 }{\sqrt {\dfrac{0.1971}{566} } }$

$z = \dfrac {-0.043 }{\sqrt {3.48233216*10^{-4} } }$

$z = \dfrac {-0.043 }{0.01866} }$

z = −2.30

C. P-value

P-value = P(Z < z)

P-value = P(Z< -2.30)

By using the P-value method and the normal distribution as an approximation to the binomial distribution.

from the table of standard normal distribution

move left until the first column is reached. Note the value as –2.0

move upward until the top row is reached. Note the value as 0.30

find the probability value as 0.010724 by the intersection of the row and column values gives the area to the left of

z = -2.30

P- value = 2P(z ≤ -2.30)

P-value = 2 × 0.01072

P - value = 0.02144

Decision Rule: Since the p-value is lesser than the level of significance; then we reject the null hypothesis.

Conclusion: We accept the alternative hypothesis and conclude that under the same circumstances the proportion of offspring peas will be yellow is not equal to 0.27

8 0

3 years ago

Because of safety considerations, in May 2003 the Federal Aviation Administration (FAA) changed its guidelines for how small com

Rasek [7]

Answer:

a) The 95% confidence interval for the mean summer weight (including carry-on luggage) of Frontier Airlines passengers is between 179 and 187 pounds. This means that we are 95% sure that the mean summer weight of all Frontier Airlines passengers is between these two values.

b)

The 95% confidence interval for the mean winter weight (including carry-on luggage) of Frontier Airlines passengers is between 185.4 pounds and 194.6 pounds. This means that we are 95% sure that the mean winter weight of all Frontier Airlines passengers is between these two values.

c) They are respected, as the upper bound of both intervals is below the new FAA recommendations.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve these questions.

Question a:

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 100 - 1 = 99

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 99 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$ . So we have T = 1.9842

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 1.9842\frac{20}{\sqrt{100}} = 4$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 183 - 4 = 179 pounds.

The upper end of the interval is the sample mean added to M. So it is 183 + 4 = 187 pounds.

The 95% confidence interval for the mean summer weight (including carry-on luggage) of Frontier Airlines passengers is between 179 and 187 pounds. This means that we are 95% sure that the mean summer weight of all Frontier Airlines passengers is between these two values.

Question b:

Critical value is the same(same sample size and confidence level).

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 1.9842\frac{23}{\sqrt{100}} = 4.6$

The lower end of the interval is the sample mean subtracted by M. So it is 190 - 4.6 = 185.4 pounds.

The upper end of the interval is the sample mean added to M. So it is 190 + 4.6 = 194.6 pounds.

The 95% confidence interval for the mean winter weight (including carry-on luggage) of Frontier Airlines passengers is between 185.4 pounds and 194.6 pounds. This means that we are 95% sure that the mean winter weight of all Frontier Airlines passengers is between these two values.

c. The new FAA recommendations are 190 pounds for summer and 195 pounds for winter. Comment on these recommendations in light of the confidence interval estimates from Parts (a) and (b).

They are respected, as the upper bound of both intervals is below the new FAA recommendations.

7 0

3 years ago

Determine the value of x. 1) 14.75 2)15.25 3)11.92 4)18.56

Leto [7]

Opposite angles are supplementary, the sum is 180 degrees.
6x+10+81.5=180
Combine like terms:
6x+91.5=180
Subtract 91.5 to both sides:
6x=88.5
Divide 6 to both sides:
x=14.75
The answer is 1) 14.75
Hope this helped!

3 0

2 years ago