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pentagon [3]
4 years ago
12

A coffee machine can be adjusted to deliver any fixed number of ounces of coffee. if the machine has a standard deviation in del

ivery equal to 0.4 ounce, what should be the mean setting so that an 8-ounce cup will overflow only 0.5% of the time?
Mathematics
1 answer:
zysi [14]4 years ago
6 0
The mean will be calculates as follows; from the formula of z-score we shall have:
z=(x-μ)/σ
the z-score associated with 0.5% is 0
thus
0=
(8-μ)/0.4=0
solving for x we get:
μ=9 

Answer: μ=9

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Two companies, A and B, make express delivery for small-item packages in a city. Company A charges a flat fee of RM70 per packag
bazaltina [42]

Using the <u>normal probability distribution and the central limit theorem</u>, it is found that:

a) There is a 0.1922 = 19.22% probability that Company B would charge more than Company A to deliver a small-item package.

b) The mean amount charged is of RM 51.5, with a standard deviation of 21.35.

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.

By the Central Limit Theorem:

  • When a <u>fixed constant k</u> multiplies a variable, the mean is k\mu and the standard deviation is k\sigma
  • When two normal variables are subtracted, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.

In this question, b is needed to solve a, so I am going to place the solution to item b first.

Item b:

Flat fee of RM20(not considered for the standard deviation), plus a variable fee of RM3.5, hence:

\mu_B = 20 + 3.5(9) = 51.5

\sigma = 6.1(3.5) = 21.35

The mean amount charged is of RM 51.5, with a standard deviation of 21.35.

Item a:

This probability is P(B - A) > 0. For the distribution, we have that:

\mu_{B-A} = \mu_B - \mu_A = 51.5 - 70 = -18.5

Since A has a constant fee, it's standard deviation is 0, hence:

\sigma_{B-A} = \sqrt{\sigma_A^2 + \sigma_B^2} = \sqrt{21.35^2} = 21.35

The probability is <u>1 subtracted by the p-value of Z when X = 0</u>, so:

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

Z = \frac{0 + 18.5}{21.35}

Z = 0.87

Z = 0.87 has a p-value of 0.8078.

1 - 0.8078 = 0.1922.

0.1922 = 19.22% probability that Company B would charge more than Company A to deliver a small-item package.

A similar problem is given at brainly.com/question/25403659

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3 years ago
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