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Novosadov [1.4K]
3 years ago
9

A pizza chain records how long it takes customers to receive their delivery orders. Suppose the distribution of these delivery t

imes is strongly skewed to the right with a mean of 30 minutes and a standard deviation of 10 minutes. Management plans on calculating the mean delivery time from a random sample of 25 orders. We can assume independence between orders in the sample.
What is the probability that the mean delivery time from the sample of 25 orders xˉ is farther than 2 minutes from the population mean?
Mathematics
1 answer:
SCORPION-xisa [38]3 years ago
5 0

Answer:

The probability that the mean delivery time from the sample of 25 orders xˉ is farther than 2 minutes from the population mean cannot be calculated.

Step-by-step explanation:

As given in the question statement, the distribution of delivery times is strongly skewed to the right. The population distribution is skewed to right. Too much skewed distribution can cause the statistical model to work ineffectively and affects its performance. The probability can also not be calculated because the sample size is too small. Small sample size affects the results and makes them less reliable because it results in a higher variability and likelihood of skewing the results.

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PLZ HELP ME! Whoever gets it right will be marked brainiest
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Answer:

Probability that Hannah only has to buy 3 or less boxes before getting a prize is 0.784

Step-by-step explanation:

Given, 40% of cereal boxes contain a prize

⇒probability of getting a prize on opening a box, P(A)=0.4

where A is the event of getting a prize on opening a cereal box

and probability of not getting a prize on opening a box, P(A')=1-P(A)=0.6

where A' is the event of not getting a prize on opening a cereal box

This problem needs to be divided into 3 situation:

  • Case 1, Where Hannah gets prize when she buys the first box:

Let K be the event of Hannah winning the prize on buying the first box.

⇒P(K)=P(A)=0.4

  • Case 2, Where Hannah gets prize when she buys the second box:

I<u>n this event Hannah should not get the prize in first box but should get the prize on buying the second box</u>

Let L be the event of Hannah winning the prize on buying the second box

So, P(L)=P(A')·P(A)

           =(0.6)·(0.4)

           =0.24

  • Case 3,Where Hannah gets prize when she buys the third box:

<u>In this event Hannah should not get the prize in first and second box but should get the prize on buying the third box</u>

Let L be the event of Hannah winning the prize on buying the third box

So, P(L)=P(A')·P(A')·P(A)

           =(0.6)·(0.6)·(0.4)

           =0.144

Let N be the event of Hannah winning the prize on buying 3 or less boxes before getting a prize

⇒N=K∪L∪M

Now, Required probability is P(N)=P(K∪L∪M)=P(K)+P(L)+P(M) [As events K,L and M are independent and disjoint events]

⇒P(N)=0.4+0.24+0.144

         =0.784

7 0
3 years ago
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