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2 years ago

5

A recent study reported that 1.5 percent of flights are canceled by major air carriers. Consider a simulation with 50 trials des

igned to estimate the number of canceled flights from a random sample of size 100, where the probability of success, a canceled flight, is 0.015.

1 answer:

Yuliya22 [10]2 years ago

0 0

Step-by-step explanation:

<em>"Of the following dotplots, which best represents the possible results from the simulation described?"</em>

The sample size is 100, and the probability of success is 0.015, so the expected value is 1.5. Meaning we would expect a dotplot with most of the dots at 1 and 2.

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Leto [7]

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4 0

2 years ago

The sum of two numbers is −138. Their difference is −2. Find the numbers.

Natasha_Volkova [10]

Answer:

The Answer is: The first number is -70 and the second number is -68.

Step-by-step explanation:

Let f = first number and s = second number.

The sum of both is -138:

f + s = -138

The difference is -2

f - s = -2

Solve for f:

f = -2 + s

f = s - 2

Substitute:

s - 2 + s = -138

2s = -136

s = -68, the second number

Solve for f:

f = s - 2

f = -68 - 2 = -70, the first number

Proof:

f + s = -138

-68 + (-70) = -138

-138 = -138

Then:

f - s = -2

-70 - (-68) = -2

-70 + 68 = -2

-2 = -2

Hope this helps! Have an Awesome Day!! :-)

5 0

2 years ago

The mean annual salary for intermediate level executives is about $74000 per year with a standard deviation of $2500. A random s

lidiya [134]

Answer:

11.51% probability that the mean annual salary of the sample is between $71000 and $73500

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 normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 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 $\mu$ 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.

In this question, we have that:

$\mu = 74000, \sigma = 2500, n = 36, s = \frac{2500}{\sqrt{36}} = 416.67$

What is the probability that the mean annual salary of the sample is between $71000 and $73500?

This is the pvalue of Z when X = 73500 subtracted by the pvalue of Z when X = 71000. So

X = 73500

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

By the Central Limit Theorem

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

$Z = \frac{73500 - 74000}{416.67}$

$Z = -1.2$

$Z = -1.2$ has a pvalue of 0.1151

X = 71000

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

$Z = \frac{71000 - 74000}{416.67}$

$Z = -7.2$

$Z = -7.2$ has a pvalue of 0.

0.1151 - 0 = 0.1151

11.51% probability that the mean annual salary of the sample is between $71000 and $73500

8 0

2 years ago

The lifespans of meerkats in a particular zoo are normally distributed. The average meerkat lives 10.410.410, point, 4 years; th

kati45 [8]

Answer:

99.85%

Step-by-step explanation:

The lifespans of meerkats in a particular zoo are normally distributed. The average meerkat lives 10.4 years; the standard deviation is 1.9 years.

Use the empirical rule (68-95-99.7%) to estimate the probability of a meerkat living less than 16.1 years.

Solution:

The empirical rule states that for a normal distribution most of the data fall within three standard deviations (σ) of the mean (µ). That is 68% of the data falls within the first standard deviation (µ ± σ), 95% falls within the first two standard deviations (µ ± 2σ), and 99.7% falls within the first three standard deviations (µ ± 3σ).

Therefore:

68% falls within (10.4 ± 1.9). 68% falls within 8.5 years to 12.3 years

95% falls within (10.4 ± 2*1.9). 95% falls within 6.6 years to 14.2 years

99.7% falls within (10.4 ± 3*1.9). 68% falls within 4.7 years to 16.1 years

Probability of a meerkat living less than 16.1 years = 100% - (100% - 99.7%)/2 = 100% - 0.15% = 99.85%

7 0

1 year ago

Read 2 more answers

Of the 500 sample households in the previous exercise, 7 had three or more large-screen TVs. (a) The percentage of households in

likoan [24]

Answer:

a) The percentage of households in the town with three or more largescreen TVs is estimated as :

The best estimation for the population proportion is :

$\hat p=\frac{7}{500}=0.014$

And that represent the 1.4%.

b) And the 95% confidence interval would be given (0.00370;0.0243).

And the % would be between 0.37% and 2.43%.

Step-by-step explanation:

Data given and notation

n=500 represent the random sample taken

X=7 represent the households with three or more large-screen TVs

$\hat p=\frac{7}{500}=0.014$ estimated proportion of households with three or more large-screen TVs

$\alpha=0.05$ represent the significance level (no given, but is assumed)

z would represent the statistic (variable of interest)

p= population proportion of households with three or more large-screen TVs

Part a

The percentage of households in the town with three or more largescreen TVs is estimated as :

The best estimation for the population proportion is :

$\hat p=\frac{7}{500}=0.014$

And that represent the 1.4%.

Part b

Yes is possible. We hav that $np>10$ and $n(1-p)>10$ so we have the assumption of normality to find the interval.

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$0.014 - 1.96 \sqrt{\frac{0.014(1-0.014)}{500}}=0.00370$

$0.014 + 1.96 \sqrt{\frac{0.014(1-0.014)}{500}}=0.0243$

And the 95% confidence interval would be given (0.00370;0.0243).

And the % would be between 0.37% and 2.43%.

4 0

2 years ago