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Volgvan
3 years ago
5

Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.68

and standard deviation 0.81. (a) If a random sample of 25 specimens is selected, what is the probability that the sample average sediment density is at most 3.00? Between 2.68 and 3.00? (Round your answers to four decimal places.) at most 3.00 between 2.68 and 3.00 (b) How large a sample size would be required to ensure that the probability in part (a) is at least 0.99? (Round your answer up to the nearest whole number.) specimens
Mathematics
1 answer:
mel-nik [20]3 years ago
5 0

Answer:

a) 0.9761; 0.4761; b) 35

Step-by-step explanation:

We use the z score formula

z=\frac{\bar{X}-\mu}{\sigma \div \sqrt{n}}

Our mean, μ, is 2.68 and our standard deviation, σ, is 0.81.  The sample size, n, is 25.

We want to find P(X ≤ 3.00):

z = (3.00-2.68)/(0.81÷√25) = 0.32/(0.81÷5) = 0.32/0.162 = 1.98

Using a z table, we see that the area under the curve to the left of this is 0.9761.

We now want to find P(2.68 ≤ X ≤ 3.00):

z = (2.68-2.68)/0.162 = 0/0.162 = 0

z = (3.00-2.68)/0.162 = 0.32/0.162 = 1.98

The area under the curve to the right of z = 0.00 is 0.5.  The area under the curve to the right of z = 1.98 is 0.9761.  This makes the area between them

0.9761-0.5000 = 0.4761.

For part b,

We use the same information, except we are now finding the sample size.  We want the probability to be 0.99.  Looking in the z table, we see that the closest number to this is 0.9901, which corresponds to a z score of 2.33:

2.33 = (3.00-2.68)/(0.81÷√n)

2.33 = 0.32/(0.81÷√n)

2.33 = 0.32×(√n/0.81)

2.33 = 0.32√n/0.81

Multiply both sides by 0.81:

0.81(2.33) = (0.32√n/0.81)(0.81)

1.8873 = 0.32√n

Divide both sides by 0.32:

1.8873/0.32 = 0.32√n/0.32

5.8978 = √x

Square both sides:

(5.8978)² = (√x)²

34.78 = x

This rounds to 35.

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Choice C: approximately 121 green beans will be 13 centimeters or shorter.

Step-by-step explanation:

What's the probability that a green bean from this sale is shorter than 13 centimeters?

Let the length of a green bean be X centimeters.

X follows a normal distribution with

  • mean \mu = 11.2 and
  • standard deviation \sigma = 2.1.

In other words,

X\sim \text{N}(11.2, 2.1^{2}),

and the probability in question is X \le 13.

Z-score table approach:

Find the z-score of this measurement:

\displaystyle z= \frac{x-\mu}{\sigma} = \frac{13-11.2}{2.1} = 0.857143. Closest to 0.86.

Look up the z-score in a table. Keep in mind that entries on a typical z-score table gives the probability of the left tail, which is the chance that Z will be less than or equal to the z-score in question. (In case the question is asking for the probability that Z is greater than the z-score, subtract the value from table from 1.)

P(X\le 13) = P(Z \le 0.857143) \approx 0.8051.

"Technology" Approach

Depending on the manufacturer, the steps generally include:

  • Locate the cumulative probability function (cdf) for normal distributions.
  • Enter the lower and upper bound. The lower bound shall be a very negative number such as -10^{9}. For the upper bound, enter 13
  • Enter the mean and standard deviation (or variance if required).
  • Evaluate.

For example, on a Texas Instruments TI-84, evaluating \text{normalcdf})(-1\text{E}99,\;13,\;11.2,\;2.1 ) gives 0.804317.

As a result,

P(X\le 13) = 0.804317.

Number of green beans that are shorter than 13 centimeters:

Assume that the length of green beans for sale are independent of each other. The probability that each green bean is shorter than 13 centimeters is constant. As a result, the number of green beans out of 150 that are shorter than 13 centimeters follow a binomial distribution.

  • Number of trials n: 150.
  • Probability of success p: 0.804317.

Let Y be the number of green beans out of this 150 that are shorter than 13 centimeters. Y\sim\text{B}(150,0.804317).

The expected value of a binomial random variable is the product of the number of trials and the probability of success on each trial. In other words,

E(Y) = n\cdot p = 150 \times 0.804317 = 120.648\approx 121

The expected number of green beans out of this 150 that are shorter than 13 centimeters will thus be approximately 121.

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