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A machine is designed to fill 16-ounce bottles of shampoo. When the machine is working properly, the amount poured into the bottles follows a normal distribution with mean 16.05 ounces with a standard deviation of .2005 ounces. If four bottles are randomly selected each hour and the number of ounces in each bottle is measured, then 95% of the means calculated should occur in what interval? Hint: the standard deviation rule says that 95% of the observations are within how many standard deviations away from the mean? Round answers to four decimal places. ( , )

Respuesta :

Answer:

(15.8495,16.2505)  

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 16.05 ounces

Standard Deviation, σ = 0.2005 ounces

We are given that the distribution of amount poured into the bottles is a bell shaped distribution that is a normal distribution.

Empirical rule:

  • It states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ).
  1. It shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).

Confidence interval:

[tex]\mu \pm 2\frac{\sigma}{\sqrt{n}}[/tex]

Putting the values, we get,

[tex]16.05 \pm 2(\frac{0.2005}{\sqrt{4}} ) = 16.05 \pm 0.2005 = (15.8495,16.2505)[/tex]

Using the normal distribution, it is found that 95% of the means calculated should occur in (15.8495, 16.2505).

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

  • 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.
  • The Central Limit Theorem states that for the sampling distribution of sample means of size n, the standard deviation is [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
  • The Empirical Rule states that 95% of the measures are within 2 standard deviations of the mean, that is, between Z = -2 and Z = 2.

In this problem:

  • Mean of 16.05 ounces, thus [tex]\mu = 16.05[/tex]
  • Standard deviation of 0.2005 ounces, thus [tex]\sigma = 0.2005[/tex]
  • Samples of size 4, thus [tex]n = 4, s = \frac{0.2005}{\sqrt{4}} = 0.10025[/tex].

95% between Z = -2 and Z = 2, thus:

Z = -2:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]-2 = \frac{X - 16.05}{0.10025}[/tex]

[tex]X - 16.05 = -2(0.10025)[/tex]

[tex]X = 15.8495[/tex]

Z = 2:

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]2 = \frac{X - 16.05}{0.10025}[/tex]

[tex]X - 16.05 = 2(0.10025)[/tex]

[tex]X = 16.2505[/tex]

Interval (15.8495, 16.2505).

A similar problem is given at https://brainly.com/question/13448290

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