Tyre pressures on a certain type of car independently follow a normal distribution with mean 1.9 bars and standard deviation 0.15 bars.
Find the probability that all four tyres on a car of this type have pressures between 1.82 bars and 1.92 bars.
Solution
First, standardize the values 1.82 and 1.92 using the formula for the z-score:
\(z_1 = \frac{1.92 - 1.9}{0.15} = 0.1333\)
\(z_2 = \frac{1.82 - 1.9}{0.15} = -0.5333\)
Next, find the probability that a single tyre has a pressure between 1.82 and 1.92 bars:
\(\text{area} = \Phi(0.1333) - \Phi(-0.5333)\)
Using standard normal distribution tables or a calculator:
\(\Phi(0.1333) = 0.5529\)
\(\Phi(-0.5333) = 1 - \Phi(0.5333) = 1 - 0.7029 = 0.2971\)
Thus, the probability for one tyre is:
\(0.5529 - 0.2971 = 0.256\)
Since the tyres are independent, the probability that all four tyres have pressures between 1.82 and 1.92 bars is:
\((0.256)^4 = 0.0043\)
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