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mel-nik [20]
3 years ago
13

At a certain airport, 65% of the flights arrive on time. A sample of 10 flights is studied. Find the probability that all 10 of

the flights were on time.
Mathematics
1 answer:
babunello [35]3 years ago
5 0

Answer:

So, the probability is P=0.0135.

Step-by-step explanation:

We know that at a certain airport, 65% of the flights arrive on time.

We get that p=0.65 and q=1-0.65=0.35.

We have 10 flights, so n=10.

We calculate the probability that all 10 of the flights were on time, so k=10.

We use the formulu:

\boxed{P(X=k)=C^n_k \cdot p^k\cdot q^{n-k}}

we get:

P(X=10)=C^{10}_{10}\cdot 0.65^{10}\cdot 0.35^0\\\\P(X=10)=1\cdot 0.0135\cdot 1\\\\P(X=10)=0.0135\\

So, the probability is P=0.0135.

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Step-by-step explanation:

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Perimeter = 1/2 (3.14 x 2) + 2

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5 0
3 years ago
In an examination of purchasing patterns of shoppers, a sample of 16 shoppers revealed that they spent, on average, $54 per hour
Lesechka [4]

Answer:

54-1.64\frac{21}{\sqrt{16}}=45.39    

54 +1.64\frac{21}{\sqrt{16}}=62.61    

So on this case the 90% confidence interval would be given by (45.39;62.61)    

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X=54 represent the sample mean  

\mu population mean (variable of interest)

\sigma=21 represent the sample standard deviation

n=16 represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

Since the Confidence is 0.90 or 90%, the value of \alpha=1-0.9=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that z_{\alpha/2}=1.64

Now we have everything in order to replace into formula (1):

54-1.64\frac{21}{\sqrt{16}}=45.39    

54 +1.64\frac{21}{\sqrt{16}}=62.61    

So on this case the 90% confidence interval would be given by (45.39;62.61)    

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