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Naya [18.7K]
3 years ago
15

A battery has a lifetime that is approximately normally distributed with a mean of 600 hours and a standard deviation of 50 hour

s. Suppose you purchased one of these batteries and it only lasted for 400 hours. Which of the following conclusions is the most appropriate given this information?
A. Your battery is likely defective since such poor performance is extremely unlikely.
B. Your battery is definitely below average since it is below the mean.
C. There is a 50-50 chance of a battery having a worse than average performance, so the performance of your battery is not surprising.
Mathematics
1 answer:
miss Akunina [59]3 years ago
3 0

Answer:

A. Your battery is likely defective since such poor performance is extremely unlikely.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

If X is more than two standard deviations from the mean, it is considered an unlikely outcome.

In this question, we have that:

\mu = 600, \sigma = 50

Which of the following conclusions is the most appropriate given this information?

Lasted 400 hours, so X = 400.

Z = \frac{X - \mu}{\sigma}

Z = \frac{400 - 600}{50}

Z = -4

4 standard deviations from the mean, so unlikely.

So the correct answer is:

A. Your battery is likely defective since such poor performance is extremely unlikely.

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Simplify the expression.<br><br> 8k - 2(4 - 3k)
DIA [1.3K]

Answer:

14k-8

Step-by-step explanation:

8k - 2(4-3k)

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8k - 8  + 6k

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In a data distribution, the first quartile, the median and the means are 30.8, 48.5 and 42.0
kolbaska11 [484]

Answer:

Q_3 = 56.45 --- The third quartile

Var = 370.18 -- Variance

Step-by-step explanation:

Given

Q_1  =  30.8 -- First quartile

Q_2 =  48.5 --- Median

\bar x = 42 --- Mean

Skp = -0.38 --- Coefficient of skewness

Solving (9): The third quartile Q_3

This is calculated from

Skp =  \frac{Q_1   + Q_3  - 2Q_2}{Q_3 - Q_1}

So, we have:

-0.38 =  \frac{30.8 + Q_3- 2*48.5}{Q_3 - 30.8}

Cross Multiply

-0.38 (Q_3 - 30.8)=  30.8 + Q_3- 2*48.5

Open bracket

-0.38Q_3 + 11.704=  30.8 + Q_3- 97.0

Collect like terms

-0.38Q_3 -Q_3=  30.8 - 97.0- 11.704

-1.38Q_3=  -77.904

Divide both sides -1.38

Q_3 = \frac{-77.904}{-1.38}

Q_3 = 56.45 --- approximated

Solving (b): The variance

First, calculate the standard deviation from:

3IQR  = 4SD

IQR= Q_3 - Q_1

So:

3IQR  = 4SD

3(Q_3 - Q_1) = 4SD

Make SD the subject

SD = \frac{3}{4}(Q_3 - Q_1)

SD = \frac{3}{4}(56.45 - 30.8)

SD = \frac{3}{4}*25.65

SD = \frac{3*25.65}{4}

SD = \frac{76.95}{4}

SD = 19.24

So, the variance is:

Var = SD^2

Var = 19.24^2

Var = 370.18

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3 years ago
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butalik [34]

We can tell that the whole L shaped angle is 90 degrees because it has the little box in the corner. Now, the 90 degree angle is split into two parts with that line coming from the box. Those two parts of the whole 90 degree angle add up to 90 degrees.

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Now, let's solve for v. Combine like terms.

90 = 13v - 1

Add 1 to both sides.

91 = 13v

Divide both sides by 13

7 = v

5 0
3 years ago
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