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Elis [28]
3 years ago
14

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Mathematics
1 answer:
ehidna [41]3 years ago
6 0

Answer:  1) 0.10

               2) 0.60

               3) 0.20

               4) 0.10

<u>Step-by-step explanation:</u>

The total frequency is 20+120+40+20 = 200.  This means they ran the experiment 200 times.  The probability distribution is calculated by the satisfactory number of outcomes (frequency) divided by the total number of experiments/outcomes (total frequency):

\begin{array}{c|c||lc}\underline{x}&\underline{f}&\underline{f\div 200}&\underline{\text{Probability Distribution}}\\1&20&20\div200=&0.10\\2&120&120\div 200=&0.60\\3&40&40\div 200=&0.20\\4&20&20\div 200=&0.10\end{array}\right]

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Step-by-step explanation:

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And we have;

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Step-by-step explanation:

Assuming this question: The delivery times for all food orders at a fast-food restaurant during the lunch hour are normally distributed with a mean of 14.7 minutes and a standard deviation of 3.7 minutes. Let R be the mean delivery time for a random sample of 40 orders at this restaurant. Calculate the mean and standard deviation of \bar X Round your answers to two decimal places.

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the delivery times of a population, and for this case we know the distribution for X is given by:

X \sim N(14.7,3.7)  

Where \mu=14.7 and \sigma=3.7

Since the distribution of X is normal then we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And we have;

\mu_{\bar X}= 14.70

\sigma_{\bar X} =\frac{3.7}{\sqrt{40}}= 0.59

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b

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sequnce 105 865

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