Ask question

Ask question

All categories

2 years ago

13

The commute time to work in the U.S. Has a bell shaped distribution with a population mean of 24.4 minutes and a population stan

dard deviation of 6.5 minutes. What proportion of the population has a commute time: a) between 11.4 minutes and 37.4 minutes? (1 pt) b) less than 11.4 minutes? (1 pt) c) greater than 37.4 minutes? (1 pt) 3. Using the parameters provided in the previous question, Q2, determine the z-score that corresponds to a commute time of 15 minutes. (2 pts)

1 answer:

Schach [20]2 years ago

7 0

Answer:

Q1 a) 0.9545

b) 0.02275

c) 0.97725

Q2 z is approximately equal to -1.466

Step-by-step explanation:

Q1 The given information are;

The mean time to commute to work = 24.4 minutes

The standard deviation = 6.5 minutes

a) The z-score for 11.4 is given as follows;

$Z=\dfrac{x-\mu }{\sigma }$

Where;

x = Observed value 11.4

μ = The mean = 24.4 minutes

σ = The standard deviation = 6.5 minutes

$Z=\dfrac{11.4-24.4 }{6.5 } = -2$

The z-score for 37.4 is given as follows;

$Z=\dfrac{37.4-24.4 }{6.5 } = 2$

-2 < z < 2, which gives, from the z-score table;

The probability of commute time to be between 11.4 minutes and 37.4 minutes = 0.97725 - 0.02275 = 0.9545

b) From the z-score table, the probability that the commute time to be less than 11.4 minutes = The probability at z = -2 = 0.02275

c) From the z-score table, the probability that the commute time to be greater than 37.4 minutes = The probability at z = 2 = 0.97725

Q2 The the z-score that corresponds to a commute time of 15 minutes is given as follows;

$Z=\dfrac{x-\mu }{\sigma }$

$Z=\dfrac{15-24.4 }{6.5 } = -\dfrac{94}{65} \approx -1.466$

You might be interested in

The California State University (CSU) system consists of 23 campuses, from San Diego State in the south to Humboldt State near t

miv72 [106K]

Answer:

For this case since since they not want to differentiate between the campuses, so is good to use a simple random sampling or SRS, that is a procedure in order to select a sample of size n from a population of size N known, and each element of the population have the sample probability of being selected $p=\frac{1}{N}$

So then the administrator can select a random sample of n students from all the campuses from the CSU University and obtain the distance from hometwon to campuses for the selected sample and then calculate the average and use this to inference.

Other possibility is use a systematic random sampling for example they can define a random number k and they can select 1 student each k individuals and then create a random sample of size n and calculate the average for this in order to do inference.

The convenience sampling is not too useful since is not a probabilistic method.

Step-by-step explanation:

For this case we can clasify this study as an enumerative study and inferential since they want to identify the average distance between the hometowns of students and their campuses.

For this case the sampling frame represent all the 23 campuses, and is known for the researcher,

For this case since since they not want to differentiate between the campuses, so is good to use a simple random sampling or SRS, that is a procedure in order to select a sample of size n from a population of size N known, and each element of the population have the sample probability of being selected $p=\frac{1}{N}$

So then the administrator can select a random sample of n students from all the campuses from the CSU University and obtain the distance from hometwon to campuses for the selected sample and then calculate the average and use this to inference.

Other possibility is use a systematic random sampling for example they can define a random number k and they can select 1 student each k individuals and then create a random sample of size n and calculate the average for this in order to do inference.

The convenience sampling is not too useful since is not a probabilistic method.

6 0

2 years ago

A company produces precision 1 meter (1000 mm) rulers. the actual distribution of lengths of the rulers produced by this company

pogonyaev

Let X be the length of rulers. X follows Normal distribution with mean μ and standard deviation σ = 0.02. A sample of size n=4 is drawn from population and sample mean m=1000. We have to find 90% confidence interval for population mean μ.

Here population standard deviation is given to us so we will use z confidence interval for mean. It is given as

(Sample mean - Margin of error , Sample mean + Margin of error)

Where margin of error for z confidence interval is

ME = σ $\frac{z_{\alpha/2}}{\sqrt{n}}$

Where σ = population standard deviation = 0.02

n= sample size =4

$z_{\alpha/2}$ = Critical z score value for given confidence level

We have to find here 90% confidence interval so confidence level c= 0.9

α = 1 -c =1-0.9 = 0.1

$z_{\alpha/2}$ = $z_{0.1/2}$ = $z_{0.05}$

Here we have to find z score value such that area below it is -z is 0.05 and above z is 0.05

Using excel function to find z score value

=NORM.S.INV(0.05) = -1.645

For calculating confidence interval we consider positive z score value which is 1.645

So the margin of error is

ME = $\frac{0.02 * 1.645}{\sqrt{4}}$

ME = 0.01645

90% confidence interval for mean is

(Sample mean - ME, Sample mean+ME)

1000- 0.01645, 1000+0.01645

(1000.01645, 999.9836)

Rounding interval upto 3 decimal places

(1000.016, 999.984)

90% confidence interval for population mean is (1000.016, 999.984)

7 0

2 years ago

the jellybean jar has a radius of 4.8 cm and a height of 11.8 cm. what would be a reasonable lower limit for the number of jelly

azamat

Answer:

$\large\boxed{\large\boxed{100}}$

Explanation:

This question comes with four answer choices:

10,000
100,000
100
1

You just need to find a rough estimate of the volume of the jar and the volume of one jelly bean. This is called magnitude order.

The <em>jellybean ja</em>r is cylindrical, Thus, its volume is $\pi \times radius^2\times height$

To find the order of magnitude, you just use the numbers rounded to one signficant figure: round π to 3, the radius to 5, cm, and the height to 10:

$Volume\approx 3cm\times (5cm)^2\times10cm=750cm^3\approx1,000cm^3$

The order of magnitude for the radius of a jellybean is 1 cm. And the order of magnitude of the volume of a sphere with a radius of 1 cm is the cube of the diameter (2cm):

$(2cm)^3=8cm^3\approx10cm^3$

Hence, a reasonable lower limit for the number of jellybeans in the jar is:

$1,000cm^3/10cm^3=100$

7 0

2 years ago

A student prepares 150.0 mL of 1.40 M HCl using 35.0 mL of a stock solution. What is the concentration of the stock solution? Us

Gwar [14]

Your answer would be c.6.00 M

5 0

2 years ago

Read 2 more answers

The ratio of length to width in a rectangle is 3 to 1. If the perimeter of the rectangle is 128 feet, what is the length of the

Serga [27]

Answer: B

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers