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Norma-Jean [14]

4 years ago

7

Among fatal plane crashes that occurred during the past 5050 years, 699699 were due to pilot error, 8888 were due to other hum

an error, 6060 were due to weather, 156156 were due to mechanical problems, and 164164 were due to sabotage. Construct the relative frequency distribution. What is the most serious threat to aviation safety, and can anything be done about it

1 answer:

lapo4ka [179]4 years ago

6 0

Answer: Crashes due to pilot error is the most serious threat to aviation safety as it has highest relative frequency.

Step-by-step explanation:

Since we have given that

Number of crashes due to pilot error = 699

Number of crashes due to other human error = 88

Number of crashes due to mechanical problems = 156

Number of crashes due to sabotage = 164

Total number of crashes = 699+88+156+164 = 1167

So, Relative frequency distribution :

Pilot error = $\dfrac{699}{1167}\times 100=59.89\%$

Human error = $\dfrac{88}{1167}\times 100=7.54\%$

Mechanical problems = $\dfrac{156}{1167}\times 100=13.36\%$

Sabtoge = $\dfrac{164}{1167}\times 100=14.05\%$

Crashes due to pilot error is the most serious threat to aviation safety as it has highest relative frequency.

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In 2002, the mean age of an inmate on death row was 40.7 years with a standard deviation of 9.6 years according to the U.S. Depa

marissa [1.9K]

Answer:

The <em>95% confidence interval</em> for the current mean age of death-row inmates is between 42.23 years and 35.57 years.

Step-by-step explanation:

The <em>confidence interval</em> of the mean is given by the next formula:

$\\ \overline{x} \pm z_{1-\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$ [1]

We already know (according to the U.S. Department of Justice):

The (population) standard deviation for this case (mean age of an inmate on death row) has a standard deviation of 9.6 years ( $\\ \sigma = 9.6$ years).
The number of observations for the sample taken is $\\ n = 32$ .
The sample mean, $\\ \overline{x} = 38.9$ years.

For $\\ z_{1-\frac{\alpha}{2}}$ , we have that $\\ \alpha = 0.05$ . That is, the <em>level of significance</em> $\\ \alpha$ is 1 - 0.95 = 0.05. In this case, then, we have that the <em>z-score</em> corresponding to this case is:

$\\ z_{1-\frac{\alpha}{2}} = z_{1-\frac{0.05}{2}} = z_{1-0.025} = z_{0.975}$

Consulting a cumulative <em>standard normal table</em>, available on the Internet or in Statistics books, to find the z-score associated to the probability of, $\\ P(z$ , we have that $\\ z = 1.96$ .

Notice that we supposed that the sample is from a population that follows a <em>normal distribution</em>. However, we also have a value for n > 30, and we already know that for this result the sampling distribution for the sample means follows, approximately, a normal distribution with mean, $\\ \mu$ , and standard deviation, $\\ \sigma_{\overline{x}} = \frac{\sigma}{\sqrt{n}}$ .

Having all this information, we can proceed to answer the question.

Constructing the 95% confidence interval for the current mean age of death-row inmates

To construct the 95% confidence interval, we already know that this interval is given by [1]:

$\\ \overline{x} \pm z_{1-\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$

That is, we have:

$\\ \overline{x} = 38.9$ years.

$\\ z_{1-\frac{\alpha}{2}} = 1.96$

$\\ \sigma = 9.6$ years.

$\\ n = 32$

Then

$\\ 38.9 \pm 1.96*\frac{9.6}{\sqrt{32}}$

$\\ 38.9 \pm 1.96*\frac{9.6}{5.656854}$

$\\ 38.9 \pm 1.96*1.697056$

$\\ 38.9 \pm 3.326229$

Therefore, the Upper and Lower limits of the interval are:

Upper limit:

$\\ 38.9 + 3.326229$

$\\ 42.226229 \approx 42.23$ years.

Lower limit:

$\\ 38.9 - 3.326229$

$\\ 35.573771 \approx 35.57$ years.

In sum, the 95% confidence interval for the current mean age of death-row inmates is between 42.23 years and 35.57 years.

Notice that the "mean age of an inmate on death row was 40.7 years in 2002", and this value is between the limits of the 95% confidence interval obtained. So, according to the random sample under study, it seems that this mean age has not changed.

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We want to know how many brownmies did Fred and his family eat together.

We will call to the total of the brownies by 1. On this case, after Fred ate 1/3 of the brownies, he will have:

$1-\frac{1}{3}=\frac{3-1}{3}=\frac{2}{3}$

This means that he has left 2/3 of the brownies. After his family ate 1/6 of the brownies:

$\frac{2}{3}-\frac{1}{6}=\frac{4}{6}-\frac{1}{6}=\frac{3}{6}=\frac{1}{2}$

This means they will have left 1/2 of the tray of brownies, and that they ate half of it.

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1 year ago