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Hatshy [7]
3 years ago
5

Carmen is a professor at a local university. In collecting data on her Introduction to Business course for a year, she wants to

calculate the z-score for a student who scores a 83 on the final exam. The mean and the standard deviation scores on the exam are 72 and 5, respectively. Calculate the z-score.
Mathematics
1 answer:
AVprozaik [17]3 years ago
3 0

Answer:

\boxed {\boxed {\sf z=2.2}}

Step-by-step explanation:

The z-score tells us how many standard deviations a value is away from the mean.

The formula is:

z=\frac{x- \overline{x}}{s}

where x is the value, \overline {x} is the mean, and s is the standard deviation.

We want to find the z-score for the score of an 83, the mean is 72, and the standard deviation is 5.

x=83 \\\overline {x}= 72 \\s=5

Substitute the values into the formula.

z=\frac{83-72}{5}

Solve the numerator.

z=\frac{11}{5}

Divide.

z=2.2

The z-score is 2.2, which means the score of 83 is 2.2 standard deviations above the mean.

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Answer:

100% probability that the sample mean weight of these 100 bags is less than 18.6 ounces

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central 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 question, we have that:

\mu = 18.5, \sigma = 0.2, n = 100, s = \frac{0.2}{\sqrt{100}} = 0.02

What is the probability that the sample mean weight of these 100 bags is less than 18.6 ounces

This is the pvalue of Z when X = 18.6. So

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

By the Central Limit Theorem

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

Z = \frac{18.6 - 18.5}{0.02}

Z = 5 has a pvalue of 1

100% probability that the sample mean weight of these 100 bags is less than 18.6 ounces

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Answer:

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Step-by-step explanation:

9.6 divided by 3/4 = 0.8, so 8. Correct me if I'm wrong ya'll.

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