It has been reported that 48% of people text and drive. A random sample of 60 drivers is drawn. Find the probability that the proportion of people in the sample who text and drive is less than 0.493 . Write only a number as your answer. Round to 4 decimal places (for example 0.3748). Do not write as a percentage.
Solution
Given that,
p = 48% = 0.48
1 - p = 1 - 0.48 = 0.52
n = 60
_{} = p = 0.48
_{} = [p ( 1 - p ) / n] = [(0.48 * 0.52 ) / 60 ] = 0.0645
P( < 0.493)
= P[( - _{} ) / _{} < ( 0.493 - 0.48 ) / 0.0645 ]
= P(z < 0.20)
Using z table
= 0.5793
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