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Bess [88]
2 years ago
12

What is the z-score with a confidence level of 95% when finding the margin of error for the mean of a normally distributed popul

ation from a sample? 0.99 1.65 1.96 2.58
Mathematics
2 answers:
Georgia [21]2 years ago
8 0

Answer:

1.96

Step-by-step explanation:

its 1.96 or c on edge 2020

PSYCHO15rus [73]2 years ago
5 0

The value of z-score with a confidence level of 95% in finding margin of error for the mean of a normally distributed population from a sample is 1.96.

<h3>What is normally distributed data?</h3>

Normally distributed data is the distribution of probability which is symmetric about the mean.

  • The mean of the data is the average value of the given data.
  • The standard deviation of the data is the half of the difference of the highest value and mean of the data set.

The confidence level is 95% when finding the margin of error for the mean of a normally distributed population from a sample.

The critical z score value for the confidence level of 95% is 1.96 and +1.96 standard deviation.

Thus, the value of z-score with a confidence level of 95% in finding margin of error for the mean of a normally distributed population from a sample is 1.96.

Learn more about the normally distributed data here;

brainly.com/question/6587992

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The amount of potato chips an 18-ounce bag contains follows a normal distribution with a mean of 18.5 ounces and a standard devi
bekas [8.4K]

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

7 0
3 years ago
Select the best answer for the question
artcher [175]

Answer:

C

HOWEVER: "Less Than" is usually depicted as a dashed line at the border. and there MUST be an OR between these two equations to include all of the blue area. An "and" there would exclude the area between the edge lines.

Step-by-step explanation:

2x + y < 10

       y < -2x + 10

3x + 2y < 12

       2y < -3x + 12

         y < -1.5x + 6

3 0
3 years ago
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