Question

# It has been reported that 48% of people text and drive. A random sample of 60...

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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