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According to the National Center for Education Statistics, 69% of Texas students are eligible to receive...

According to the National Center for Education Statistics, 69% of Texas students are eligible to receive free or reduced-price lunches. Suppose you randomly choose 285 Texas students. Find the probability that no more than 73% of them are eligible to receive free or reduced-price lunches.

Expert Solution

Solution

Given that,

p = 0.69

1 - p = 1 - 0.69 = 0.31

n = 285

= p = 0.69

= $\sqrt$ p ( 1 - p ) / n

= $\sqrt$ (0.69 * 0.31) / 285 = 0.0274

P( 0.73) =

= P(( - ) / (0.73 - 0.69) / 0.0274)

= P(z 1.46)

= 0.9279

Probability = 0.9279

milcah answered 1 year ago

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