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Question 11 of 40

Suppose the probability density function of the length of computer cables is [tex]\( f(x)=0.1 \)[/tex] from 1200 to 1210 millimeters.

a) Determine the mean and standard deviation of the cable length.

Mean [tex]\( = \ \square \)[/tex] millimeters

Standard deviation [tex]\( = \ \square \)[/tex] 2.89 millimeters (Round the answer to 2 decimal places.)

b) If the length specifications are [tex]\( 1195\ \textless \ x\ \textless \ 1205 \)[/tex], what proportion of cables is within specifications?

[tex]\( \square \)[/tex] (Round the answer to 1 decimal place.)

Sagot :

GinnyAnswer
Let's tackle this problem step-by-step.

### Part (a) - Mean and Standard Deviation

#### Mean:
The mean (or expected value) of a uniform distribution [tex]\( U(a, b) \)[/tex] is calculated as:
[tex]\[ \text{Mean} = \frac{a + b}{2} \][/tex]

Given bounds for the uniform distribution are 1200 and 1210 millimeters.
[tex]\[ \text{Mean} = \frac{1200 + 1210}{2} = \frac{2410}{2} = 1205 \][/tex]

So, the mean length of the cable is:
[tex]\[ \text{Mean} = 1205 \, \text{millimeters} \][/tex]

#### Standard Deviation:
The standard deviation of a uniform distribution [tex]\( U(a, b) \)[/tex] is given by:
[tex]\[ \text{Standard deviation} = \frac{b - a}{\sqrt{12}} \][/tex]

Using the given bounds:
[tex]\[ \text{Standard deviation} = \frac{1210 - 1200}{\sqrt{12}} = \frac{10}{\sqrt{12}} \approx 2.89 \][/tex]

So, the standard deviation of the cable length, rounded to two decimal places, is:
[tex]\[ \text{Standard deviation} = 2.89 \, \text{millimeters} \][/tex]

### Part (b) - Proportion Within Specifications

To find the proportion of cables within the specification limits of [tex]\( 1195 < x < 1205 \)[/tex], we standardize these limits into z-scores and then use the cumulative distribution function (CDF) of the standard normal distribution.

#### Z-Scores:
A z-score is calculated as:
[tex]\[ z = \frac{x - \text{mean}}{\text{standard deviation}} \][/tex]

The specification limits are 1195 and 1205 millimeters.

##### Lower specification limit (1195):
[tex]\[ z_{\text{lower}} = \frac{1195 - 1205}{2.89} = \frac{-10}{2.89} \approx -3.46 \][/tex]

##### Upper specification limit (1205):
[tex]\[ z_{\text{upper}} = \frac{1205 - 1205}{2.89} = \frac{0}{2.89} = 0 \][/tex]

#### Proportion within specifications:
We look up these z-scores in the standard normal distribution table or use the CDF of the normal distribution.

- The CDF for [tex]\( z_{\text{lower}} \approx -3.46 \)[/tex] is almost 0.
- The CDF for [tex]\( z_{\text{upper}} = 0 \)[/tex] is 0.5.

Thus, the proportion within specifications is:
[tex]\[ P(1195 < x < 1205) = \text{CDF}(0) - \text{CDF}(-3.46) = 0.5 - 0 \approx 0.5 \][/tex]

So, the proportion of cables within the specified limits, rounded to one decimal place, is:
[tex]\[ \text{Proportion within specifications} = 0.5 \][/tex]

### Final Answers:
(a) The mean length of the cable is [tex]\( 1205 \)[/tex] millimeters, and the standard deviation is [tex]\( 2.89 \)[/tex] millimeters.
(b) The proportion of cables within the specified limits is [tex]\( 0.5 \)[/tex].
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