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kvv77 [185]
2 years ago
8

Or

Mathematics
1 answer:
Neporo4naja [7]2 years ago
8 0

The probability that a single can will contain 11.6 ounces or less of soda is 0.2843

<h3>Probability that a can contains 11.6 ounces or less</h3>

The given parameters are:

x = 11.6

Mean = 12

Standard deviation = 0.7

Calculate the z value using:

z = \frac{x - \bar x}{\sigma}

This gives

z = \frac{11.6-12}{0.7}

z = -0.57

The probability is then calculated as:

P(x ≤ 11.6) = P(z ≤ -0.57)

Using the z table of probabilities, we have:

P(x ≤ 11.6) = 0.2843

<h3>Probability that a pack contains 11.6 ounces or less</h3>

In (a), the probability that a can contains 11.6 ounces or less is 0.2843

The probability that all cans in a pack contains 11.6 ounces or less is

P(6) = 0.2843^6

P(6) = 0.00053

<h3>Probability that a case contains 11.6 ounces or less</h3>

In (a), the probability that a can contains 11.6 ounces or less is 0.2843

The probability that all cans in a case contains 11.6 ounces or less is

P(36) = 0.2843^36

P(36) ≈ 0

<h3>Draw three normal distributions</h3>

See attachment for the normal distributions

<h3>The happening on the graph</h3>

The summary of the graph is that, as the sample size increases the probability decreases

Read more about probability at:

brainly.com/question/11234923

#SPJ1

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Suppose that the total cost of flatiron TVs is given by the function TC(Q) = 500Q + 300 dollars, where Q is the number of TVs pr
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The value of TC(10) is 5300 Dollars

Given function is TC(Q) = 500Q +300 dollars

We need to calculate the value for TC (10)

As we know that the equation given is TC(Q) = 500Q + 300 dollars

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Therefore, The value of Q is 10 for TC(10)

Substituting the value of Q

in the total cost of flatiron TVs given by function TC(Q) = 500Q + 300 dollars That Q= 100,

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2 years ago
A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than ex
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Answer:

a

 The  90% confidence interval that  estimate the true proportion of students who receive financial aid is

     0.533  <  p <  0.64

b

   n = 1789

Step-by-step explanation:

Considering question a

From the question we are told that

      The sample size is  n = 200

      The number of student that receives financial aid is k = 118

Generally the sample proportion is  

      \^ p = \frac{k}{n}

=>   \^ p = \frac{118}{200}

=>   \^ p = 0.59

From the question we are told the confidence level is  90% , hence the level of significance is    

      \alpha = (100 - 90 ) \%

=>   \alpha = 0.10

Generally from the normal distribution table the critical value  of \frac{\alpha }{2}  is  

   Z_{\frac{\alpha }{2} } =  1.645

Generally the margin of error is mathematically represented as  

     E =  Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }

 =>E =  1.645 * \sqrt{\frac{0.59 (1- 0.59)}{200} }

=>  E = 0.057

Generally 90% confidence interval is mathematically represented as  

      \^ p -E <  p <  \^ p +E

  =>  0.533  <  p <  0.64  

Considering question b

From the question we are told that

    The margin of error  is  E = 0.03

From the question we are told the confidence level is  99% , hence the level of significance is    

      \alpha = (100 - 99 ) \%

=>   \alpha = 0.01

Generally from the normal distribution table the critical value  of   is  

   Z_{\frac{\alpha }{2} } = 2.58

Generally the sample size is mathematically represented as      

        [\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p )

=>      n = [\frac{2.58}{0.03} ]^2 * 0.59 (1 - 0.59 )

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