The plane is vertical. What best describes the shape of the cross section?

A savings and loan association needs information concerning the checking account balances of its local customers. A random sampl

Marina86 [1]

Answer:

The 99% confidence interval for the true mean checking account balance for local customers is ($439.29, $888.99).

Step-by-step explanation:

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

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 = 14 - 1 = 13

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 13 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.99}{2} = 0.995$ . So we have T = 3.0123

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 3.0123\frac{279.29}{\sqrt{14}} = 224.85$

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 664.14 - 224.85 = $439.29

The upper end of the interval is the sample mean added to M. So it is 664.14 + 224.85 = $888.99.

The 99% confidence interval for the true mean checking account balance for local customers is ($439.29, $888.99).

4 0

2 years ago

A dragon likes to eat red peppers and green peppers because they help him breathe fire. If the dragon eats 2 red peppers, he can

N76 [4]

it's 45

Step-by-step explanation:

Okay, so if 2 peppers is 25 meters long, you just divide 25 by 2 to get how many peppers the dragon breathes as 1 pepper. So then you get 12.5, And if you do that with the green peppers, you get 1.5. So, the dragon ate 3 green peppers, so it's 12.5 x 3 , aka. 12.5+12.5+12.5. That's 37.5, and if you do 1.5x5 for the green peppers, it's 7.5, And now you add 37.5+7.5, and you get 45.

4 0

1 year ago

0<br> x+y- 2x +284 +18120<br> The answer

Ganezh [65]

Answer:

doint know yup

Step-by-step explanation:

8 0

2 years ago

Write 0.5% as a decimal.

Pavlova-9 [17]

0.5% = 0.005
0.5% = 1/200

6 0

2 years ago

Read 2 more answers

The number of boots that 25 students had in their homes in Florida were recorded below:

Doss [256]

Answer:

A) 8

B) Significantly (increased)

C) Not affected

D) Median

E) Significantly (increased)

F) Not affected

G) Interquartile range

Step-by-step explanation:

A)

When we have a certain set of data, we call "outlier" a value in this set of data that is significantly far from all the other values.

In this problem, we see that we have the following set of data:

0,0,0,0,0,0,0,1,1,1,1,2,2,2,2,2,2,2,3,3,3,3,4,5,8

We see that all the values between 0 and 5 are relatively close to each other, being only 1 number distant from any other value.

On the other hand, the last value "8" is 3 numbers distant from the other highest value ("5"), so the outliar is 8.

B)

The mean of a certain set of value indicates the "average" value of the dataset. It is given by:

$\bar x = \frac{\sum x_i}{N}$

where

$\sum x_i$ is the sum of all the values in the dataset

N is the number of values in the dataset (in this case, 25)

An outlier usually affects a lot the mean, because it changes the value of the sum $\sum x_i$ by quite a lot.

In fact in this problem, we have

$\sum x_i = 47$

N = 25

So the mean is

$\bar x=\frac{47}{25}=1.88$

Instead, without the outliar "8",

$\sum x_i = 39\\N=24$

And the mean would be

$\bar x=\frac{39}{24}=1.63$

So we see that the outlier increases the mean significantly.

C)

The median of a dataset is the "central value" of the dataset: it means it is the value for which 50% of the values of the dataset are above the median and 50% of the values of the dataset are below the median.

In this problem, we have 25 values in the dataset: so the median is the 13th values of the ordered dataset (because if we take the 13th value, there are 12 data below it, and 12 data above it). The 13th value here is a "2", so the median is 2.

We see also that the outlier does not affect the median.

In fact, if we had a "5" instead of a "8" (so, not an outlier), the median of this dataset would still be a 2.

D)

The center of a distribution is described by two different quantities depending on the conditions:

- If the distribution is symmetrical and has no tails/outliers, then the mean value is a good value that can be used to describe the center of the distribution.

- On the other hand, if the distribution is not very symmetrical and has tails/outliers, the median is a better estimator of the center of the distribution, because the mean is very affected by the presence of outliers, while the median is not.

So in this case, since the distribution is not symmetrical and has an outlier, it is better to use the median to describe the center of the distribution.

E)

The standard deviation of a distribution is a quantity used to describe the spread of the distribution. It is calculated as:

$\sigma = \sqrt{\frac{\sum (x_i-\bar x)^2}{N}}$

where

$\sum (x_i-\bar x)^2$ is the sum of the residuals

N is the number of values in the dataset (in this case, 25)

$\bar x$ is the mean value of the dataset

The presence of an outlier in the distribution affects significantly the standard deviation, because it changes significantly the value of the sum $\sum (x_i-\bar x)^2$ .

In fact, in this problem,

$\sum (x_i-\bar x)^2=84.64$

So the standard deviation is

$\sigma = \sqrt{\frac{84.64}{25}}=1.84$

Instead, if we remove the outlier from the dataset,

$\sum (x_i-\bar x)^2=45.63$

So the standard deviation is

$\sigma = \sqrt{\frac{45.63}{24}}=1.38$

So, the standard deviation is much higher when the outlier is included.

F)

The interquartile range (IQ) of a dataset is the difference between the 3rd quartile and the 1st quartile:

$IQ=Q_3-Q_1$

Where:

$Q_3$ = 3rd quartile is the value of the dataset for which 25% of the values are above $Q_3$ , and 75% of the values are below it

$Q_1$ = 1st quartile is the value of the dataset for which 75% of the values are above $Q_1$ , and 25% of the values are below it

In this dataset, the 1st and 3rd quartiles are the 7th and the 18th values, respectively (since we have a total of 25 values), so they are:

$Q_1=0\\Q_3=2$

So the interquartile range is

$IQ=2-0 = 2$

If the outlier is removed, $Q_1, Q_3$ do not change, therefore the IQ range does not change.

G)

The spread of a distribution can be described by using two different quantities:

- If the distribution is quite symmetrical and has no tails/outliers, then the standard deviation is a good value that can be used to describe the spread of the distribution.

- On the other hand, if the distribution is not very symmetrical and has tails/outliers, the interquartile range is a better estimator of the spread of the distribution, because the standard deviation is very affected by the presence of outliers, while the IQ range is not is not.

So in this case, since the distribution is not symmetrical and has an outlier, it is better to use the IQ range to describe the spread of the distribution.

5 0

2 years ago