Write an expression that is equivalent to . Then, find the quotient.
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The z-score of the speed value gives the measure of dispersion of the from
the mean observed speed.
The probability that the speed of a car is between 63 km/h and 75 km/h is
<u>0.273</u>.
The given parameters are;
The mean of the speed of cars on the highway, = 60 km/h
The standard deviation of the cars on the highway, σ = 5 km/h
Required:
The probability that the speed of a car is between 63 km/h and 75 km/h
Solution;
The z-score for a speed of 63 km/h is given as follows;
Which gives;
From the z-score table, we have;
P(x < 63) = 0.7257
The z-score for a speed of 75 km/h is given as follows;
Which gives, P(x < 75) = 0.9987
The probability that the speed of a car is between 63 km/h and 75 km/h is therefore;
P(63 < x < 75) = P(x < 75) - P(x < 63) = 0.9987 - 0.7257 = 0.273
The probability that the speed of a car is between 63 km/h and 75 km/h is
<u>0.273</u>.
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