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You own a branch of Good-Buy Electronics and have been told by the manufacturer of Stevuski...

You own a branch of Good-Buy Electronics and have been told by the manufacturer of Stevuski Televisions that only 5% of their brand of TVs die within one year. Last year, your branch sold 130 such televisions in July. One year later, 14 (about 11%) of them had been returned — dead.

(a) Assuming the 5% value quoted by the manufacturer was accurate, what is the mean number of TVs that die within one year in randomly samples of size 130? Round your answer to one decimal place.
μ =

(b) What is the standard deviation? Round your answer to one decimal place.
σ =

(c) You had 14 TVs die out of 130 (about 11%). With respect to the mean and standard deviation found in parts (a) & (b) respectively, what is the z-score for this many dead TVs? Round your answer to two decimal places.
z =

(d) Assuming the 5% rate is accurate, would 14 dead TVs in a sample of 130 TVs be considered unusual?

Yes, that is an unusual number of dead TVs.No, that is not unusual.    

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Answer #2

(a) Assuming the 5% value quoted by the manufacturer was accurate, the mean number of TVs that die within one year in randomly sampled of size 130 can be calculated using the formula:

μ = n * p

where n is the sample size and p is the probability of a TV dying within one year.

μ = 130 * 0.05 = 6.5

The mean number of TVs that die within one year is 6.5.

(b) The standard deviation can be calculated using the formula:

σ = sqrt(n * p * (1 - p))

σ = sqrt(130 * 0.05 * (1 - 0.05)) ≈ 2.39

The standard deviation is approximately 2.39.

(c) To calculate the z-score, we can use the formula:

z = (x - μ) / σ

where x is the observed value, μ is the mean, and σ is the standard deviation.

In this case, x = 14, μ = 6.5, and σ ≈ 2.39.

z = (14 - 6.5) / 2.39 ≈ 3.14

The z-score is approximately 3.14.

(d) To determine if 14 dead TVs in a sample of 130 TVs is considered unusual, we can compare the z-score to a certain threshold. A common threshold is 2 standard deviations from the mean, which corresponds to a z-score of ±2.

In this case, the z-score of 3.14 is greater than 2, indicating that the observed number of dead TVs is more than 2 standard deviations away from the mean. Therefore, 14 dead TVs in a sample of 130 TVs would be considered unusual.


answered by: mervetokaz
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