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 0.498.

Solutions

Expert Solution

Solution:

Given that,

n = 60

= 48% =0.48

1 - = 1 - 0.48 = 0.52

= = 0.48

= $\sqrt$ ( 1 - ) / n

= $\sqrt$ 0.48 * 0.52 / 60

= 0.0645

p ( < 0.498 )

p ( - / ) < ( 0.498 - 0.48 / 0.0645 )

p ( z < 0.018 / 0.0645 )

Using z table

p ( z < 0.28 )

= 0.6103

Probability = 0.6103

orchestra answered 1 year ago

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