Find the interquartile range for this set of data.

An airline, believing that 5% of passengers fail to show for flights, overbooks (sells more tickets than there are seats). Suppo

Fittoniya [83]

Answer:

Q1. 13 passengers

Q2. 0.1756 (approx. 0.18)

Step-by-step explanation:

Q1. 267 seats are available on the plane

5% is expected to fail to show up

Hence, no of passengers expected not to show up = 267 * 0.05

= 13.35 (approx 13 passengers)

Q2. See working in the attachment as I had to explain it step by step.

6 0

2 years ago

What two products would you add to find 713x48

umka21 [38]

9514 1404 393

Answer:

40·713 and 8·713

Step-by-step explanation:

When this multiplication is carried out "by hand", the usual sum of partial products is ...

8·713 + 40·713

5 0

2 years ago

Read 2 more answers

The time for a visitor to read health instructions on a Web site is approximately normally distributed with a mean of 10 minutes

klio [65]

Answer:

a) The mean is 10 and the variance is 0.0625.

b) 0.6826 = 68.26% probability that the mean time of the visitors is within 15 seconds of 10 minutes.

c) 10.58 minutes.

Step-by-step explanation:

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

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

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

Normally distributed with a mean of 10 minutes and a standard deviation of 2 minutes.

This means that $\mu = 10, \sigma = 2$

Suppose 64 visitors independently view the site.

This means that $n = 64, = \frac{2}{\sqrt{64}} = 0.25$

a. The expected value and the variance of the mean time of the visitors.

Using the Central Limit Theorem, mean of 10 and variance of (0.25)^2 = 0.0625.

b. The probability that the mean time of the visitors is within 15 seconds of 10 minutes.

15 seconds = 15/60 = 0.25 minutes, so between 9.75 and 10.25 seconds, which is the p-value of Z when X = 10.25 subtracted by the p-value of Z when X = 9.75.

X = 10.25

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{10.25 - 10}{0.25}$

$Z = 1$

$Z = 1$ has a p-value of 0.8413.

X = 9.75

$Z = \frac{X - \mu}{s}$

$Z = \frac{9.75 - 10}{0.25}$

$Z = -1$

$Z = -1$ has a p-value of 0.1587.

0.8413 - 0.1587 = 0.6826.

0.6826 = 68.26% probability that the mean time of the visitors is within 15 seconds of 10 minutes.

c. The value exceeded by the mean time of the visitors with probability 0.01.

The 100 - 1 = 99th percentile, which is X when Z has a p-value of 0.99, so X when Z = 2.327.

$Z = \frac{X - \mu}{s}$

$2.327 = \frac{X - 10}{0.25}$

$X - 10 = 2.327*0.25$

$X = 10.58$

So 10.58 minutes.

6 0

2 years ago

The length of a rectangular picture is 5 inches more than three times the width. Find the dimensions of the picture if its perim

luda_lava [24]

Answer:

<h3>Length = 29 inches</h3><h3>Width = 8 inches</h3><h3 />

Step-by-step explanation:

Perimeter of a rectangle = 2l + 2w

where

l is the length of the rectangle

w is the width

From the question

length of a rectangular picture is 5 inches more than three times the width is written as

l = 5 + 3w

Now substitute this into the above equation

Perimeter = 74 inches

74 = 2(5 + 3w) + 2w

74 = 10 + 6w + 2w

8w = 74 - 10

8w = 64

Divide both sides by 8

w = 8 inches

Substitute w = 8 into l = 5 + 3w

That's

l = 5 + 3(8)

l = 5 + 24

l = 29 inches

<h3>Length = 29 inches</h3><h3>Width = 8 inches</h3>

Hope this helps you

4 0

2 years ago

Read 2 more answers

In constructing a 95 percent confidence interval, if you increase n to 4n, the width of your confidence interval will (assuming

Damm [24]

Answer:

about 50 percent of its former width.

Step-by-step explanation:

Let's assume that our parameter of interest is given by $\theta$ and in order to construct a confidence interval we can use the following formula:

$\hat \theta \pm ME(\hat \theta)$

Where $\hat \theta$ is an estimator for the parameter of interest and the margin of error is defined usually if the distribution for the parameter is normal as:

$ME = z_{\alpha} SE$

Where $z_{\alpha/2}$ is a quantile from the normal standard distribution that accumulates $\alpha/2$ of the area on each tail of the distribution. And $SE$ represent the standard error for the parameter.

If our parameter of interest is the population proportion the standard of error is given by:

$SE= \frac{\hat p (1-\hat p)}{n}$

And if our parameter of interest is the sample mean the standard error is given by:

$SE = \frac{s}{\sqrt{n}}$

As we can see the standard error for both cases assuming that the other things remain the same are function of n the sample size and we can write this as:

$SE = f(n)$

And since the margin of error is a multiple of the standard error we have that $ME = f(n)$

Now if we find the width for a confidence interval we got this:

$Width = \hat \theta + ME(\hat \theta) -[\hat \theta -ME(\hat \theta)]$

$Width = 2 ME (\hat \theta)$

And we can express this as:

$Width =2 f(n)$

And we can define the function $f(n) = \frac{1}{\sqrt{n}}$ since as we can see the margin of error and the standard error are function of the inverse square root of n. So then we have this:

$Width_i= 2 \frac{1}{\sqrt{n}}$

The subscript i is in order to say that is with the sample size n

If we increase the sample size from n to 4n now our width is:

$Width_f = 2 \frac{1}{\sqrt{4n}} =2 \frac{1}{\sqrt{4}\sqrt{n}} =\frac{2}{2} \frac{1}{\sqrt{n}} =\frac{1}{\sqrt{n}} =\frac{1}{2} Width_i$

The subscript f is in order to say that is the width for the sample size 4n.

So then as we can see the width for the sample size of 4n is the half of the wisth for the width obtained with the sample size of n. So then the best option for this case is:

about 50 percent of its former width.

7 0

2 years ago