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finlep [7]
4 years ago
8

Two types of interiors A & B are designed for Honda Accord 2018 model. Assume that the total customer review score (out of 1

00) for these two interior designs is independent and normally distributed. The review score of part A has a mean of 55 and a standard deviation of 5. The review score of part B has a mean of 60 and a standard deviation of 10. What is the probability that the review score of design B minus review score of design A is greater than 5?
Mathematics
1 answer:
Dahasolnce [82]4 years ago
3 0

Answer:

50%

Step-by-step explanation:

<u>Distribution 1</u>

Mean

\bf \bar x_1 = 55

<u>Distribution 2</u>

Mean

\bf \bar x_2 = 60

<u>Distribution 2 - Distribution 1</u>

Mean

\bf \bar x_2-\bar x_1=60-55=5

What is the probability that the review score of design B minus review score of design A is greater than 5

<em>In any normal distribution X, the probability that X is greater than its mean is 50% or 0.5, so the probability that the review score of design B minus review score of design A is greater than 5 is 50%  </em>

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Answer:

Image point → (-4, 0)

Step-by-step explanation:

Coordinate of a point → (0, 1)

Rule for the transformation has been given as,

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Answer:

6.26-2.01\frac{2.47}{\sqrt{50}}=5.56    

6.26+2.01\frac{2.47}{\sqrt{50}}=6.96    

So on this case the 95% confidence interval would be given by (5.56;6.96)

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 represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

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

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

The mean calculated for this case is \bar X=6.26

The sample deviation calculated s=2.47

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=50-1=49

Assuming a Confidence of 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,49)".And we see that t_{\alpha/2}=2.01

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

6.26-2.01\frac{2.47}{\sqrt{50}}=5.56    

6.26+2.01\frac{2.47}{\sqrt{50}}=6.96    

So on this case the 95% confidence interval would be given by (5.56;6.96)    

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Step-by-step explanation:

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