Suppose the speeds of cars along a stretch of I-40 is normally distributed with a mean of 70 mph and standard deviation of 8 mph.
What would be the speed for a car whose z-score is z=−1.6?
f a car is chosen at random on this stretch of highway, what is the probability it will be travelling at a speed below 70 mph?
If a car is chosen at random on this stretch of highway, what is the probability it will be travelling at a speed greater than or equal to 86 mph? (Give your answer to 2 decimal
If a car is chosen at random on this stretch of highway, what is the probability it will be travelling at a speed between 70 mph and 86 mph? (Give your answer to 2 decimal
What is the probability that a car chosen at random on this highway will be traveling between 65 mph and 80 mph? (Give your answer to 2 decimal places.)
Solution:-
Given that,
mean = = 70
standard deviation = = 8
a) z = -1.6
Using z-score formula,
x = z * +
x = -1.6 * 8 + 70
x = 57.2 mph
b) P(x < 70) = P[(x - ) / < (70 - 70) / 8]
= P(z < 0)
Using z table,
= 0.50
c) P(x 86) = 1 - P(x 86)
= 1 - P[(x - ) / ( 86 - 70) / 8]
= 1 - P(z 2.0)
Using z table,
= 1 - 0.98
= 0.02
d) P(70 < x < 86) = P[(70 - 70)/ 8) < (x - ) / < (86 - 70) / 8 ) ]
= P(0 < z < 2.0)
= P(z < 2.0) - P(z < 0)
Using z table,
= 0.98 - 0.50
= 0.48
e) P(65 < x < 80) = P[(65 - 70)/ 8) < (x - ) / < (80 - 70) / 8 ) ]
= P(-0.63 < z < 1.25)
= P(z < 1.25) - P(z < -0.63)
Using z table,
= 0.89 - 0.26
= 0.63
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