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choli [55]
3 years ago
12

The weights of a certain brand of candies are normally distributed with a mean weight of 0.8544 g and a standard deviation of 0.

0525 g. A sample of these candies came from a package containing 466 candies, and the package label stated that the net weight is 397.9 g. (If every package has 466 candies, the mean weight of the candies must exceed 397.9466
Required:
If 1 candy is randomly​ selected, find the probability that it weighs more than 0.8535 g. The probability is:_____
Mathematics
1 answer:
aleksandr82 [10.1K]3 years ago
8 0

Answer:

The probability is 0.508 = 50.8%.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

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.

Normally distributed with a mean weight of 0.8544 g and a standard deviation of 0.0525 g.

This means that \mu = 0.8544, \sigma = 0.0525

If 1 candy is randomly​ selected, find the probability that it weighs more than 0.8535 g.

This is 1 subtracted by the pvalue of Z when X = 0.8535. So

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

Z = \frac{0.8535 - 0.8544}{0.0525}

Z = -0.02

Z = -0.02 has a pvalue of 0.492

1 - 0.492 = 0.508

The probability is 0.508 = 50.8%.

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Read 2 more answers
Scores at a local high school on the American College Testing (ACT) college entrance exam follow the normal distribution with a
san4es73 [151]

Answer:

a) The mean is 18 and the standard deviation is 1.79.

b) The interpretation is that the standard deviation of the sample means of groups of 20 students will be of 1.79, which is the sample error, which is different from the population standard deviation.

c) 0.2005 = 20.05% probability that a sample of 20 students has a mean score of 19.5 or more.

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 normal distributions can be solved using the z-score formula.

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.

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.

Mean of 18 and a standard deviation of 8.

This means that \mu = 18, \sigma = 8

Sample of 20:

This means that n = 20, s = \frac{8}{\sqrt{20}} = 1.79

(a) Calculate the mean and standard deviation of the sampling distribution of x¯.

By the Central Limit Theorem, the mean is 18 and the standard deviation is 1.79.

(b) Interpret the standard deviation from part (a).

The interpretation is that the standard deviation of the sample means of groups of 20 students will be of 1.79, which is the sample error, which is different from the population standard deviation.

(c) Find the probability that a sample of 20 students has a mean score of 19.5 or more.

This is 1 subtracted by the pvalue of Z when X = 19.5. So

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

By the Central Limit Theorem

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

Z = \frac{19.5 - 18}{1.79}

Z = 0.84

Z = 0.84 has a pvalue of 0.7995

1 - 0.7995 = 0.2005

0.2005 = 20.05% probability that a sample of 20 students has a mean score of 19.5 or more.

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zhuklara [117]
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