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Vaselesa [24]
2 years ago
8

The sales data for January and February of a frozen yogurt shop are approximately normal. The mean daily sales for January was $

300 with a standard deviation of $20. On the 15th of January, the shop sold $310 of yogurt. The mean daily sales for February was $320 with a standard deviation of $50. On the 15th of February, the shop sold $340 of yogurt. Which month had a higher z-score for sales on the 15th, and what is the value of that z-score
Mathematics
1 answer:
VashaNatasha [74]2 years ago
3 0

Answer:

January had a higher z-score for sales on the 15th, and the value of that z-score was of 0.5.

Step-by-step explanation:

z-score:

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

January:

The mean daily sales for January was $300 with a standard deviation of $20. On the 15th of January, the shop sold $310 of yogurt. This means, respectively, that \mu = 300, \sigma = 20, X = 310. So

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

Z = \frac{310 - 300}{20}

Z = 0.5

February:

The mean daily sales for February was $320 with a standard deviation of $50. On the 15th of February, the shop sold $340 of yogurt. This means, respectively, that \mu = 320, \sigma = 50, X = 340. So

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

Z = \frac{340 - 320}{50}

Z = 0.4

January had a higher z-score for sales on the 15th, and the value of that z-score was of 0.5.

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The inside diameter of a randomly selected piston ring is a random variable with mean value 13 cm and standard deviation 0.08 cm
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a) P(12.99 ≤ X ≤ 13.01) = 0.3840

b) P(X ≥ 13.01) = 0.3075

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the cental limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 13, \sigma = 0.08

(a) Calculate P(12.99 ≤ X ≤ 13.01) when n = 16.

Here we have n = 16, s = \frac{0.08}{\sqrt{16}} = 0.02

This probability is the pvalue of Z when X = 13.01 subtracted by the pvalue of Z when X = 12.99.

X = 13.01

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

By the Central Limit Theorem

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

Z = \frac{13.01 - 13}{0.02}

Z = 0.5

Z = 0.5 has a pvalue of 0.6915

X = 12.99

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

Z = \frac{12.99 - 13}{0.02}

Z = -0.5

Z = -0.5 has a pvalue of 0.3075

0.6915 - 0.3075 = 0.3840

P(12.99 ≤ X ≤ 13.01) = 0.3840

(b) How likely is it that the sample mean diameter exceeds 13.01 when n = 25?

P(X ≥ 13.01) =

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Z = \frac{13.01 - 13}{0.02}

Z = 0.5

Z = 0.5 has a pvalue of 0.6915

1 - 0.6915 = 0.3075

P(X ≥ 13.01) = 0.3075

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