Solve the equation. 33 = p – 6.71 A. –39.71 B. –26.29 C. 39.71 D. 26.29
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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%.
There is one clock that shows the right time so we do not have to worry about the one which is always correct.
Talking about the second clock that loses a minutes in every 24 hours (or in a day), so after 60 days (since it has lost 60 minutes because it is losing 1 minute everyday) it will show 11:00 a.m when it is exactly the noon.
So this way, in total it will take days before it shows the correct noon.
Now, the third clock gains a minute every 24 hours (or in a day) , after 60 days (when it has gained 60 minutes or a complete hour) it will show 1:00 p.m when it is exactly the noon.
This way, it will take days (since it has gained a minute everyday) when it shows the correct noon.
Therefore, it will take 1440 days before all the three clocks show the correct time again.