The highway fuel economy of a 2016 Lexus RX 350 FWD 6-cylinder 3.5-L automatic 5-speed using premium fuel is a normally distributed random variable with a mean of μ = 26.50 mpg and a standard deviation of σ = 3.25 mpg.

Required:
a. What is the standard error of X and the mean from a random sample of 25 fill-ups by one driver?
b. Within what interval would you expect the sample mean to fall, with 98 percent probability?

Answers

Answer 1

Answer:

a) 0.65 mpg

b) Between 24.99 mpg and 28.01 mpg.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation, which is also called standard error, [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

[tex]\mu = 26.50, \sigma = 3.25, n = 25, s = \frac{3.25}{\sqrt{25}} = 0.65[/tex]

a. What is the standard error of X and the mean from a random sample of 25 fill-ups by one driver?

s = 0.65 mpg

b. Within what interval would you expect the sample mean to fall, with 98 percent probability?

From the: 50 - (98/2) = 1st percentile

To the: 50 + (98/2) = 99th percentile

1st percentile:

X when Z has a pvalue of 0.01. So X when Z = -2.327.

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

By the Central Limit Theorem

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

[tex]-2.327 = \frac{X - 26.50}{0.65}[/tex]

[tex]X - 26.50 = -2.327*0.65[/tex]

[tex]X = 24.99[/tex]

99th percentile:

X when Z has a pvalue of 0.99. So X when Z = 2.327.

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

[tex]2.327 = \frac{X - 26.50}{0.65}[/tex]

[tex]X - 26.50 = 2.327*0.65[/tex]

[tex]X = 28.01[/tex]

Between 24.99 mpg and 28.01 mpg.


Related Questions

Question 15 A party rental company has chairs and tables for rent. The total cost to rent 8 chairs and 3 tables is $38 . The total cost to rent 2 chairs and 5 tables is $35 . What is the cost to rent each chair and each table?

Answers

Answer:

Each table is $6 and each chair is $2.50

Step-by-step explanation:

A 12 sided die is rolled the set of equally likely outcomes is 123 456-789-10 11 and 12 find the probability of rolling a number greater than three

Answers

Answer:

6

Step-by-step explanation:

nerd physics

Perform the indicated operation.

Answers

Answer:

√75 = 5√3 and √12 = 2√3 so √75 + √12 = 5√3 + 2√3 = 7√3.

Answer:

[tex] 7\sqrt{3} [/tex]

Step-by-step explanation:

[tex] \sqrt{12} \: can \: be \: simplified \: as \: 2 \sqrt{3} \: and \: \sqrt{75} \: canbe \: simplified \: as \: 5 \sqrt{3} \\ after \: simplifying \: we \: can \: add \: them \: up \\ 2 \sqrt{3} + 5 \sqrt{3} = 7 \sqrt{3} [/tex]

The Ball Corporation's beverage can manufacturing plant in Fort Atkinson, Wisconsin, uses a metal supplier that provides metal with a known thickness standard deviation σ = .000586 mm. Assume a random sample of 59 sheets of metal resulted in an x¯ = .2905 mm. Calculate the 95 percent confidence interval for the true mean metal thickness.

Answers

Answer:

The 95 percent confidence interval for the true mean metal thickness is between 0.2903 mm and 0.2907 mm

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1-0.95}{2} = 0.025[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].

So it is z with a pvalue of [tex]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/tex]

Now, find the margin of error M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

[tex]M = 1.96\frac{0.000586}{\sqrt{59}} = 0.0002[/tex]

The lower end of the interval is the sample mean subtracted by M. So it is 0.2905 - 0.0002 = 0.2903 mm

The upper end of the interval is the sample mean added to M. So it is 0.2905 + 0.0002 = 0.2907 mm

The 95 percent confidence interval for the true mean metal thickness is between 0.2903 mm and 0.2907 mm

Please answer this correctly

Answers

Step-by-step explanation:

pnotgrt8rthan4 = 3 ÷ 7 × 100

= 42.8571428571 / 43%

CAN SOMEONE HELP ME ASAP







A. 5
B. 53‾√53
C. 10
D. 103√3

Answers

Answer:

n = 5

Step-by-step explanation:

Since this is a right triangle, we can use trig functions

tan theta = opp/ adj

tan 30 = n/ 5 sqrt(3)

5 sqrt(3) tan 30 = n

5 sqrt(3) * 1/ sqrt(3) = n

5 = n

The curvature of a plane parametric curve x = f(t), y = g(t) is $ \kappa = \dfrac{|\dot{x} \ddot{y} - \dot{y} \ddot{x}|}{[\dot{x}^2 + \dot{y}^2]^{3/2}}$ where the dots indicate derivatives with respect to t. Use the above formula to find the curvature. x = 6et cos(t), y = 6et sin(t)

Answers

Answer:

The curvature is modelled by [tex]\kappa = \frac{e^{-t}}{6\sqrt{2}}[/tex].

Step-by-step explanation:

The equation of the curvature is:

[tex]\kappa = \frac{|\dot {x}\cdot \ddot {y}-\dot{y}\cdot \ddot{x}|}{[\dot{x}^{2}+\dot{y}^{2}]^{\frac{3}{2} }}[/tex]

The parametric componentes of the curve are:

[tex]x = 6\cdot e^{t} \cdot \cos t[/tex] and [tex]y = 6\cdot e^{t}\cdot \sin t[/tex]

The first and second derivative associated to each component are determined by differentiation rules:

First derivative

[tex]\dot{x} = 6\cdot e^{t}\cdot \cos t - 6\cdot e^{t}\cdot \sin t[/tex] and [tex]\dot {y} = 6\cdot e^{t}\cdot \sin t + 6\cdot e^{t} \cdot \cos t[/tex]

[tex]\dot x = 6\cdot e^{t} \cdot (\cos t - \sin t)[/tex] and [tex]\dot {y} = 6\cdot e^{t}\cdot (\sin t + \cos t)[/tex]

Second derivative

[tex]\ddot{x} = 6\cdot e^{t}\cdot (\cos t-\sin t)+6\cdot e^{t} \cdot (-\sin t -\cos t)[/tex]

[tex]\ddot x = -12\cdot e^{t}\cdot \sin t[/tex]

[tex]\ddot {y} = 6\cdot e^{t}\cdot (\sin t + \cos t) + 6\cdot e^{t}\cdot (\cos t - \sin t)[/tex]

[tex]\ddot{y} = 12\cdot e^{t}\cdot \cos t[/tex]

Now, each term is replaced in the the curvature equation:

[tex]\kappa = \frac{|6\cdot e^{t}\cdot (\cos t - \sin t)\cdot 12\cdot e^{t}\cdot \cos t-6\cdot e^{t}\cdot (\sin t + \cos t)\cdot (-12\cdot e^{t}\cdot \sin t)|}{\left\{\left[6\cdot e^{t}\cdot (\cos t - \sin t)\right]^{2}+\right[6\cdot e^{t}\cdot (\sin t + \cos t)\left]^{2}\right\}^{\frac{3}{2}}} }[/tex]

And the resulting expression is simplified by algebraic and trigonometric means:

[tex]\kappa = \frac{72\cdot e^{2\cdot t}\cdot \cos^{2}t-72\cdot e^{2\cdot t}\cdot \sin t\cdot \cos t + 72\cdot e^{2\cdot t}\cdot \sin^{2}t+72\cdot e^{2\cdot t}\cdot \sin t \cdot \cos t}{[36\cdot e^{2\cdot t}\cdot (\cos^{2}t -2\cdot \cos t \cdot \sin t +\sin^{2}t)+36\cdot e^{2\cdot t}\cdot (\sin^{2}t+2\cdot \cos t \cdot \sin t +\cos^{2} t)]^{\frac{3}{2} }}[/tex]

[tex]\kappa = \frac{72\cdot e^{2\cdot t}}{[72\cdot e^{2\cdot t}]^{\frac{3}{2} } }[/tex]

[tex]\kappa = [72\cdot e^{2\cdot t}]^{-\frac{1}{2} }[/tex]

[tex]\kappa = 72^{-\frac{1}{2} }\cdot e^{-t}[/tex]

[tex]\kappa = \frac{e^{-t}}{6\sqrt{2}}[/tex]

The curvature is modelled by [tex]\kappa = \frac{e^{-t}}{6\sqrt{2}}[/tex].

PLEASE ANSWER FAST, THANKS! :)

Answers

Answer:

Step-by-step explanation:

k = 3 ; 2k + 2 = 2*3 + 2 = 6 + 2 = 8

k = 4;  2k + 2 = 2*4 + 2 = 8 +2 = 10

k =5; 2k + 2 = 2*5 +2 = 10+2 = 12

k=6;  2k +2 = 2*6 + 2 = 12+2 = 14

k = 7 ; 2k + 2 = 2*7 +2 = 14 +2 = 16

k = 8 ; 2k + 2 = 2*8 + 2 = 16 +2 = 18

∑ (2k + 2) = 8 + 10 + 12 + 14 + 16 + 18 = 78

Cheryl bought 3.4 pounds of coffee that cost $6.95 per pound . How many did she spend on coffee

Answers

Answer:

23.63

Step-by-step explanation:

multiply the cost by the pounds

Answer:

$23.63

Step-by-step explanation:

3.4 X 6.95 = 23.63

16. How much money will I need to have at retirement so I can withdraw $60,000 a year for 20 years from an account earning 8% compounded annually? a. How much do you need in your account at the beginning b. How much total money will you pull out of the account? c. How much of that money is interest?

Answers

Answer:

starting balance: $636,215.95total withdrawals: $1,200,000interest withdrawn: $563,784.05

Step-by-step explanation:

a) If we assume the annual withdrawals are at the beginning of the year, we can use the formula for an annuity due to compute the necessary savings.

The principal P that must be invested at rate r for n annual withdrawals of amount A is ...

  P = A(1+r)(1 -(1 +r)^-n)/r

  P = $60,000(1.08)(1 -1.08^-20)/0.08 = $636,215.95

__

b) 20 withdrawals of $60,000 each total ...

  20×$60,000 = $1,200,000

__

c) The excess over the amount deposited is interest:

  $1,200,000 -636,215.95 = $563,784.05

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