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The weekly demand for a particular automobile manufacturer follows a normal distribution with a mean of 40,000 cars and a standard deviation of 10,000. Below you will find probability and percentile calculations related to the customer purchase amounts. Use this information to answer the following questions. Probability Calculations P(Sales < 2,000,000) = 0.134, P(Sales < 2,050,000) = 0.339 P(Sales < 2,100,000) = 0.609, P(Sales < 2,150,000) = 0.834 Percentiles Calculations 1st percentile = 1,912,245, 5th percentile = 1,961,388 95th percentile = 2,198,612, 99th percentile = 2,247,755 What is the probability that this company will sell between 2.0 and 2.15 million cars next year? 0.495 1 0.609 0.50

          The weekly demand for a particular automobile manufacturer follows a normal distribution with a mean of 40,000 cars and a standard deviation of 10,000. Below you will find probability and percentile calculations related to the customer purchase amounts. Use this information to answer the following questions.

Probability Calculations
P(Sales < 2,000,000) = 0.134, P(Sales < 2,050,000) = 0.339
P(Sales < 2,100,000) = 0.609, P(Sales < 2,150,000) = 0.834

Percentiles Calculations
1st percentile = 1,912,245, 5th percentile = 1,961,388
95th percentile = 2,198,612, 99th percentile = 2,247,755

What is the probability that this company will sell between 2.0 and 2.15 million cars next year?
0.495
1
0.609
0.50
        
Show more…
The weekly demand for a particular automobile manufacturer follows a normal distribution with a mean of 40,000 cars and a standard deviation of 10,000. Below you will find probability and percentile calculations related to the customer purchase amounts. Use this information to answer the following questions.

Probability Calculations
P(Sales < 2,000,000) = 0.134, P(Sales < 2,050,000) = 0.339
P(Sales < 2,100,000) = 0.609, P(Sales < 2,150,000) = 0.834

Percentiles Calculations
1st percentile = 1,912,245, 5th percentile = 1,961,388
95th percentile = 2,198,612, 99th percentile = 2,247,755

What is the probability that this company will sell between 2.0 and 2.15 million cars next year?
0.495
1
0.609
0.50

Added by Whitney D.

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Elementary Statistics a Step by Step Approach
Elementary Statistics a Step by Step Approach
Allan G. Bluman 9th Edition
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The weekly demand for a particular automobile manufacturer follows a normal distribution with a mean of 40,000 cars and a standard deviation of 10,000. Below you will find probability and percentile calculations related to the customer purchase amounts. Use this information to answer the following questions. Probability Calculations P(Sales < 2,000,000) = 0.134, P(Sales < 2,050,000) = 0.339 P(Sales < 2,100,000) = 0.609, P(Sales < 2,150,000) = 0.834 Percentiles Calculations 1st percentile = 1,912,245, 5th percentile = 1,961,388 95th percentile = 2,198,612, 99th percentile = 2,247,755 What is the probability that this company will sell between 2.0 and 2.15 million cars next year? 0.495 1 0.609 0.50
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Transcript

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00:01 In this table we have the probability distribution for a discrete random variable x, which happens to be the number of cars sold daily for one week.
00:10 And for question one, we are asked for the probability of selling at least four cars daily.
00:16 So this is the probability that x is greater than are equal to four.
00:24 And so if we look at our table, the values, the possible values for x that are at least four are four, six, eight, and ten.
00:32 So the probability of x being any of these values is the sum of their individual probabilities.
00:41 Now first we must find the probability that x equals 4.
00:47 For any discrete probability distribution, summation of all of the probability masses must be 1.
01:16 And so if we solve for the probability that x equals 4, we get 0 .35.
01:34 So the probability that x is at least 4 is the sum of the 4 probabilities highlighted in the 0.
01:39 In yellow, which can be re -expressed as 1 minus the probability that x equals 2.
01:50 So that's 0 .8.
01:52 So the correct answer for question 1 is therefore c.
02:00 And then for question 2, we are asked how many cars the dealer ship expects to sail daily.
02:08 So this is the expected value of x...
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