Answer:
(a) We can expect 95% of the batteries to last between 33 and 45 months.
(b) We can expect 0.15% of the batteries to last less than 30 months.
(c) We can expect 16% of the batteries to last longer than 42 months.
(d) We can expect 49.85% of the batteries to last between 30 and 39 months.
(e) We can expect 15.85% of the batteries to last between 42 and 48 months.
Step-by-step explanation:
The Empirical Rule states that for a normal distribution:
- Approximately 68% of the data falls within 1 standard deviation of the mean.
- Approximately 95% of the data falls within 2 standard deviations of the mean.
- Approximately 99.7% of the data falls within 3 standard deviations of the mean.
For the True Test auto battery:
- Mean (μ) = 39 months
- Standard deviation (σ) = 3 months
As the standard deviation is 3 months, the x-values for 1, 2 and 3 standard deviations from the mean are:
[tex]\sf \mu + \sigma = 39 + 3 = 42\\\\\mu + 2\sigma = 39 + 2(3) = 45\\\\\mu + 3\sigma = 39 + 3(3) = 48\\\\\mu - \sigma = 39 - 3 = 36\\\\\mu - 2\sigma = 39 - 2(3) = 33\\\\\mu - 3\sigma = 39 - 3(3) = 30\\\\[/tex]
Now, draw a horizontal axis and a bell-shaped curve. Label the mean (39) at the center. Mark and label the points for 1, 2, and 3 standard deviations from the mean. Add the probabilities between each interval. (See attachment).
In summary:
- 68% of the batteries last between 36 and 42 months.
- 95% of the batteries last between 33 and 45 months.
- 99.7% of the batteries last between 30 and 48 months.
[tex]\dotfill[/tex]
Using the drawn normal distribution curve, we can calculate the specified probabilities, keeping in mind that the curve is symmetrical about the mean.
(a) We have already determined that 95% of the batteries last between 33 and 45 months.
(b) To determine the probability of the batteries lasting less than 30 months, subtract 99.7% from 100% and divide the result by 2:
[tex]\sf P(X < 30)=\dfrac{100\%-99.7\%}{2}\\\\\\P(X < 30)=0.15\%[/tex]
(c) To determine the probability of the batteries lasting longer than 42 months, subtract 34% from 50%:
[tex]\sf P(X > 42)=50\%-34\%\\\\P(X > 42)=16\%[/tex]
(d) To determine the probability of the batteries lasting between 30 and 39 months, divide 99.7% by 2:
[tex]\sf P(30 < X < 39)=\dfrac{99.7\%}{2}\\\\\\P(30 < X < 39)=49.85\%[/tex]
(e) To determine the probability of the batteries lasting between 42 and 48 months, subtract 34% from half of 99.7%
[tex]\sf P(42 < X < 48)=\dfrac{99.7\%}{2}-34\%\\\\\\P(42 < X < 48)=49.85\%-34\%\\\\\\P(42 < X < 48)=15.85\%[/tex]
