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IRINA_888 [86]
2 years ago
7

A breakfast cereal producer makes its most popular product by combining just raisins and flakes in each box of cereal. The amoun

ts of flakes in the boxes of this cereal are normally distributed with a mean of 370\,\text{g}370g370, start text, g, end text and a standard deviation of 24\,\text{g}24g24, start text, g, end text. The amounts of raisins are also normally distributed with a mean of 170\,\text{g}170g170, start text, g, end text and a standard deviation of 7\,\text{g}7g7, start text, g, end text.
Let T=T=T, equals the total amount of product in a randomly selected box, and assume that the amounts of flakes and raisins are independent of each other.
Find the probability that the total amount of product is less than 575\,\text{g}575g575, start text, g, end text.
You may round your answer to two decimal places.
P(T<575)\approx
Mathematics
1 answer:
yarga [219]2 years ago
5 0

Using the normal distribution, it is found that there is a 0.9192 = 91.92% probability that the total amount of product is less than 575 g.

In a normal distribution with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

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

  • It 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, which is the percentile of X.
  • When two normal variables are added, the mean is the sum of the means while the standard deviation is the square root of the sum of the variances.

In this problem, the product is composed by flakes and raisins, and we have that:

\mu_F = 370, \sigma_F = 24, \mu_R = 170, \sigma_R = 7

Hence, the distribution for the total amount of product has <u>mean and standard deviation</u> given by:

\mu = \mu_F + \mu_R = 370 + 170 = 540

\sigma = \sqrt{\sigma_F^2 + \sigma_R^2} = \sqrt{24^2 + 7^2} = 25

The probability that the total amount of product is less than 575 g is the <u>p-value of Z when X = 575</u>, hence:

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

Z = \frac{575 - 540}{25}

Z = 1.4

Z = 1.4 has a p-value of 0.9192.

0.9192 = 91.92% probability that the total amount of product is less than 575 g.

A similar problem is given at brainly.com/question/22934264

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Vector v is plotted below. What is the length of the x-component of v?
Rzqust [24]
Answer:

C. 1

Step-by-step explanation:

We can see that each box in the grid is one unit.

We count one box to the right on the horizontal axis and 4 boxes down on the vertical axes to obtain the components of vector v.

See graph in attachment.

Therefore the components of vector v is

\binom{1}{-4}.

The length of the x component is 1 unit

Hence the correct answer is C.

7 0
3 years ago
Consider the question of whether the home team wins more than half of its games in the National Basketball Association. Suppose
spayn [35]

Answer:

0.0037 = 0.37% probability that the home team would win 65% or more of its games in a simple random sample of 80 games

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 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.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

The home team therefore wins 50% of its games

This means that p = 0.5

Determine the probability that the home team would win 65% or more of its games in a simple random sample of 80 games

Sample of 80 means that n = 80 and, by the Central Limit Theorem:

\mu = p = 0.65

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.5*0.5}{80}} = 0.0559

This probability is 1 subtracted by the pvalue of Z when X = 0.65. So

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

By the Central Limit Theorem

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

Z = \frac{0.65 - 0.5}{0.0559}

Z = 2.68

Z = 2.68 has a pvalue of 0.9963

1 - 0.9963 = 0.0037

0.0037 = 0.37% probability that the home team would win 65% or more of its games in a simple random sample of 80 games

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Step-by-step explanation:

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