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The minimum injection pressure (psi) for injection molding specimens of high amylose corn was determined for eight different specimens (higher pressure corresponds to greater processing difficulty), resulting in the following observations. 14.9 12.6 17.7 14.2 12.5 11.0 9.3 8.1 (a) Determine the values of the sample mean x, sample median , and 12.5% trimmed mean xtr. (Round your answers to two decimal places.) x = psi = psi xtr = psi Compare these values. The mean is much larger than the median and trimmed mean, indicating negative skewness. All three measures of center are similar, indicating little skewness to the data set. The median is much larger than the mean and trimmed mean, indicating positive skewness. The median is much larger than the mean and trimmed mean, indicating negative skewness. The mean is much larger than the median and trimmed mean, indicating positive skewness. (b) By how much could the smallest sample observation, currently 8.1, be increased without affecting the value of the sample median? psi (c) Suppose we want the values of the sample mean and median when the observations are expressed in kilograms per square inch (ksi) rather than psi. Is it necessary to reexpress each observation in ksi, or can the values calculated in part (a) be used directly? [Hint: 1 kg = 2.2 lb.] Yes, it is necessary to reexpress each observation. No, the values obtained in part (a) can be used directly.

Respuesta :

Answer and Step-by-step explanation:

(a) To calculate the sample mean:

x = ∑xi / n

x = 12.5375

Sample median is the middle term of the set. To determine it, put the data in order: 8.1,9.3,11,12.5,12.6,14.2,14.9,17.7

There are 8 numbers in the set, so the middle term is the average of 12.5 and 12.6:

median = [tex]\frac{12.5+12.6}{2}[/tex]

median = 12.55

The 12.5% trimmed mean is the mean calculated excluding the 12.5% largest and the 12.5% smallest values form the sample and taking the average of the remaining numbers.

In this case, there are 8 numbers:

12.5% of 8 = 1

So, to find the trimmed mean:

[tex]x_{tr} = \frac{9.3+11+12.5+12.6+14.2+14.9}{6}[/tex]

[tex]x_{tr}[/tex] = 12.42

The results show that the three measures of center are similar, indicating little skewness to the data set, i.e., the graph is a bell shaped one.

(b) The smallest sample could be increased by any number lower than 12.5, since the last number affects directly the sample median.

(c) No, the values obtained in (a) can be used directly: The mean in psi can be rewritten in ksi:

(12.5375) psi . [tex]\frac{1 ksi}{2.2 psi}[/tex] = 5.7 ksi

Part(a): The required values are,

Mean=[tex]12.54[/tex]

Median=[tex]12.55[/tex]

Trimmed Mean=[tex]12.42[/tex]

Part(b):

The required value is [tex]D=4.4 psi[/tex]

Part(c):

No, the values obtained in part (a) can be used directly.

Mean and Median:

The formula of mean is,

[tex]\bar{x}=\frac{\sum {x}}{n}[/tex]

And The formula of the median is,

[tex]Median=(\frac{n+1}{2} ) th[/tex] observation.

Given data in ascending order is,

[tex]8.1,9.3,11,12.5,12.6,14.2,14.9,17.7[/tex]

Part(a):

Applying the given data into the above formula we get,

[tex]\bar{x} =\frac{100.3}{8} \\\bar{x} =12.54[/tex]

[tex]Median=\frac{12.5+12.6}{2}\\Median =12.55[/tex]

Now, calculating trimmed mean.

Remove high and low observation from the data then,

[tex]trimmed\ mean=\frac{9.3+11+12.5+12.6+14.2+14.9}{6}\\ =12.42[/tex]

Part(b):

[tex]8.1[/tex] can go up to [tex]12.5[/tex] then,

[tex]D=12.5-8.1\\D=4.4 psi[/tex]

Part(c):

No, the values obtained in part (a) can be used directly.

Learn more about mean and median:

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