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A study finds that the carapace length of an adult spider is normally distributed with a mean of [tex]14.05 \, \text{mm}[/tex] and a standard deviation of [tex]1.44 \, \text{mm}[/tex]. Let [tex]x[/tex] denote carapace length for the adult spider.

a. Sketch the distribution of the variable [tex]x[/tex]. Choose the correct graph below.
A.
B.
C.

b. Obtain the standardized version, [tex]z[/tex], of [tex]x[/tex]. Choose the correct standardized version below.
A. [tex]z = \frac{(x - 14.05)}{1.44}[/tex]
B. [tex]z = \frac{(x - 1.44)}{14.05}[/tex]
C. [tex]x = \frac{(z - 14.05)}{1.44}[/tex]
D. [tex]x = \frac{(z - 1.44)}{14.05}[/tex]

Sagot :

GinnyAnswer
Step-by-Step Solution:

### Part (a): Sketch the Distribution

In the problem, it's stated that the carapace length of an adult spider is normally distributed with:
- Mean ([tex]\(\mu\)[/tex]) = 14.05 mm
- Standard Deviation ([tex]\(\sigma\)[/tex]) = 1.44 mm

However, the rough sketch itself won't be provided in text. To sketch this distribution on a graph, you would:
1. Draw a horizontal axis labelled with carapace length values (in mm).
2. Mark the mean value at [tex]\(14.05\)[/tex] mm on this axis.
3. Spread out three to four standard deviations to the left and right of the mean. For instance:
- Mean: 14.05
- One SD (14.05 - 1.44 = 12.61, 14.05 + 1.44 = 15.49)
- Two SDs (14.05 - 2.88 = 11.17, 14.05 + 2.88 = 16.93)
4. The curve should be bell-shaped, centered at the mean, and symmetrical about the vertical line through the mean.
5. The highest point of the bell curve should be at [tex]\(14.05\)[/tex], and the tails should taper off as you move further from the mean.

You need to look at the provided choices (A, B, C) and pick the one that represents this normal distribution centered at [tex]\(14.05\)[/tex] with the appropriate spread.

### Part (b): Obtain the Standardized Version

To standardize the variable [tex]\(x\)[/tex], we need to convert it into a z-score. The formula for calculating the z-score is based on the mean and standard deviation of the normal distribution:

The formula for the z-score is:
[tex]\[ z = \frac{x - \mu}{\sigma} \][/tex]

Using the given values:
- [tex]\(\mu\)[/tex] (mean) = 14.05 mm
- [tex]\(\sigma\)[/tex] (standard deviation) = 1.44 mm

So, the correct formula for standardizing [tex]\(x\)[/tex] to become [tex]\(z\)[/tex] is:
[tex]\[ z = \frac{x - 14.05}{1.44} \][/tex]

Upon reviewing the given options, we can see:
- Option A: [tex]\( z = \frac{(x - 14.05)}{1.44} \)[/tex]
- Option B: [tex]\( z = \frac{(x - 1.44)}{14.05} \)[/tex]
- Option C: [tex]\( x = \frac{(z - 14.05)}{1.44} \)[/tex]
- Option D: [tex]\( x = \frac{(z - 1.44)}{14.05} \)[/tex]

The correct standardized version, which aligns with our computed formula, is:
Option A: [tex]\( z = \frac{(x - 14.05)}{1.44} \)[/tex]

Thus, the standardized version (z) of [tex]\(x\)[/tex] should be formulated as:
[tex]\[ z = \frac{x - 14.05}{1.44} \][/tex]

So, the correct answer for part (b) is Option A.
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