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Vlad [161]
3 years ago
13

The manager of a local soft drink bottling company believes that a new beverage- dispensing machine, which is designed to dispen

se 7 ounces, in fact dispenses anywhere between 6.5 and 7.5 ounces.
If the amount dispensed is represented by a uniformly distributed random variable x:
(a) Find the mean and standard deviation of x.
(b) Find the probability that x is at least 7 ounces.
(c) Find the probability that x is between 6.5 and 7.25 ounces.
Mathematics
1 answer:
Alexxandr [17]3 years ago
7 0

Answer:

a) The mean is 7 ounces and the standard deviation is of 0.29 ounces.

b) 50% probability that x is at least 7 ounces.

c) 75% probability that x is between 6.5 and 7.25 ounces.

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The probability that we find a value X at least x is given by the following formula.

P(X \geq x) = 1 - \frac{x - a}{b-a}

The probability that we find a value X between c and d is given by the following formula.

P(c \leq X \leq d) = \frac{d - c}{b-a}

The mean of the uniform distribution is:

M = \frac{a + b}{2}

The standard deviation of the uniform distribution is:

S = \sqrt{\frac{(b-a)^{2}}{12}}

Anywhere between 6.5 and 7.5 ounces.

This means that a = 6.5, b = 7.5

(a) Find the mean and standard deviation of x.

Mean

M = \frac{6.5 + 7.5}{2} = 7

Standard deviation

S = \sqrt{\frac{{7.5-6.5)^{2}}{12}} = 0.29

The mean is 7 ounces and the standard deviation is of 0.29 ounces.

(b) Find the probability that x is at least 7 ounces.

P(X \geq 7) = 1 - \frac{7 - 6.5}{7.5 - 6.5} = 0.5

50% probability that x is at least 7 ounces.

(c) Find the probability that x is between 6.5 and 7.25 ounces.

P(6.5 \leq X \leq 7.25) = \frac{7.25 - 6.5}{7.5 - 6.5} = 0.75

75% probability that x is between 6.5 and 7.25 ounces.

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Answer:

  1. 96
  2. 6.9
  3. 25%
  4. 8%
  5. 79
  6. 33.8

Step-by-step explanation:

Here is the complete question: Find each number. Round to the nearest tenth if necessary.

  1. 80% of 120.
  2. 115% of 6
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  4. 3 is what percent of 36?
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1) 80% of 120

= \frac{80}{100} \times 120= 8\times 12

= 96

∴ 80% of 120= 96

2) 115\%\ of\ 6

=\frac{115}{100}\times 6= 1.15\times 6

∴ 115\%\ of\ 6 = 6.9

3) What percent of 128 is 32?

= \frac{32}{128} \times 100

= 25\%

4) 3 is what percent of 36?

= \frac{3}{36} \times 100

= 8\%

5) 23.7 is 30% of what number?

Let the number be "x".

⇒ \frac{30}{100} \times x= 23.7

Cross multiplying

⇒ x= \frac{23.7\times 100}{30}

∴ x= 79.

6) 71/2% of what is 12?

Lets assume the required number be "x".

⇒ \frac{71}{2} \times 100\times x= 12

cross multiplying both side.

⇒x= 33.8

∴ 71/2% of 33.8 is 12

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