Question

the lifetime of a car batteries is normally distributed with mean 1 year and standard deviation 1.5 months
a)calculate the probabilty that a randomly selected car battery will last 10 to 11 months
b)calculate the minimum lifetime of the 30% longest lasting car batteries
c) by improing the quality of the car batteries, the lifetime is to be increased such that only 2.5% of the car batteries will fail before 10 months.Find the new mean lifetime of the car batteries, Assume the standard deviation remains unchanged,

Answer 1

Answer on Question #79585 – Math – Statistics and Probability

The lifetime of a car batteries is normally distributed with mean 1 year and standard deviation 1.5 months.

Question

a) calculate the probability that a randomly selected car battery will last 10 to 11 months.

Solution

$$z = \frac {x - \mu}{\sigma};$$ $$z _ {1} = \frac {x _ {1} - \mu}{\sigma} = \frac {10 - 12}{1.5} = -1.33$$ $$z _ {2} = \frac {11 - 12}{1.5} = -0.67$$ $$p \left(x _ {1} < x < x _ {2}\right) = p \left(z _ {1} < z < z _ {2}\right) = p \left(z < z _ {2}\right) - p \left(z < z _ {1}\right);$$ $$p \left(z < z _ {1}\right) = 0.0912$$ $$p \left(z < z _ {2}\right) = 0.2525$$ $$p \left(x _ {1} < x < x _ {2}\right) = 0.2525 - 0.0912 = 0.1613.$$

Question

b) calculate the minimum lifetime of the $30\%$ longest lasting car batteries

Solution

$$Z_{0.3} = 0.52;$$ $$x = \mu + z\sigma = 12 + 0.52 \times 1.5 = 12.78.$$

Question

c) by improving the quality of the car batteries, the lifetime is to be increased such that only 2.5% of the car batteries will fail before 10 months. Find the new mean lifetime of the car batteries, Assume the standard deviation remains unchanged,

Solution

$$Z_{0.025} = -1.96$$ $$x = \mu + z\sigma; \text{ hence } \mu = x - z\sigma = 10 - (-1.96) \times 1.5 = 12.94.$$

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