The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widget weights have a mean of 46 o
1 answer:
Answer:
(a) 68% of the widget weights lie between <u>43 ounces</u> and <u>49 ounces</u>.
(b) The percentage of the widget weights lie between 43 and 87 ounces is 15.87%.
(c) The percentage of the widget weights lie below 76 is 100%.
Step-by-step explanation:
Let <em>X</em> = weight of widgets manufactured by Acme Company.
The distribution of the random variable <em>X</em> is, N (<em>μ </em>= 46, <em>σ</em>²<em> </em>=<em> </em>3²).
According to the Empirical Rule in a normal distribution with mean <em>µ</em> and standard deviation <em>σ</em>, nearly all the data will fall within 3 standard deviations of the mean. The empirical rule can be broken into three parts:
- 68% data falls within 1 standard deviation of the mean. That is P (µ - σ ≤ X ≤ µ + σ) = 0.68.
- 95% data falls within 2 standard deviations of the mean. That is P (µ - 2σ ≤ X ≤ µ + 2σ) = 0.95.
- 99.7% data falls within 3 standard deviations of the mean. That is P (µ - 3σ ≤ X ≤ µ + 3σ) = 0.997.
(a)
According to the Empirical rule, 68% data falls within 1 standard deviation of the mean.
P (µ - σ ≤ X ≤ µ + σ) = 0.68.
Compute the upper and lower values as follows:
<em>µ</em> - <em>σ</em> = 46 - 3 = 43 ounces
<em>µ</em> + <em>σ</em> = 46 + 3 = 49 ounces
Thus, 68% of the widget weights lie between <u>43 ounces</u> and <u>49 ounces</u>.
(b)
Compute the probability of the widget weights lie between 43 and 87 ounces as follows:
*Use a <em>z</em>-table.
The percentage is, 0.1587 × 100 = 15.87%.
Thus, the percentage of the widget weights lie between 43 and 87 ounces is 15.87%.
(c)
Compute the probability of the widget weights lie below 76 as follows:
*Use a <em>z</em>-table.
The percentage is, 1 × 100 = 100%.
Thus, the percentage of the widget weights lie below 76 is 100%.
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1. if the highest degree (highest exponent) in the numerator is bigger than that of the denominator, then there won't be any horizontal asymptote.
2. if the highest degree in the denominator is bigger, then the horizontal symptote would be y = 0.
3. if they have the same highest degree, then we just get the quotient of their coefficient.
Now, going back to our function, we have
From this we can see that the highest degree in the numerator is 1 (from 2x) and 2 (from x²) for the denominator. Clearly, it shows that its denominator has a higher degree. And from our discussion, we can conclude that the horizontal asymptote would be y = 0.
Answer: y = 0
1. if the highest degree (highest exponent) in the numerator is bigger than that of the denominator, then there won't be any horizontal asymptote.
2. if the highest degree in the denominator is bigger, then the horizontal symptote would be y = 0.
3. if they have the same highest degree, then we just get the quotient of their coefficient.
Now, going back to our function, we have
From this we can see that the highest degree in the numerator is 1 (from 2x) and 2 (from x²) for the denominator. Clearly, it shows that its denominator has a higher degree. And from our discussion, we can conclude that the horizontal asymptote would be y = 0.
Answer: y = 0