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You have a sample proportion (p-hat) that is is normally distributed, where p = 0.6 and...

You have a sample proportion (p-hat) that is is normally distributed, where p = 0.6 and n = 32. The likelihood that the sample proportion is greater than .65 is :

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Answer #1

Solution

Given that,

=  [p( 1 - p ) / n] = [(0.6 * 0.4) / 32 ] = 0.0866

P( > .65) = 1 - P( < .65)

= 1 - P(( - ) / < (.65 - 0.6) / 0.0866)

= 1 - P(z < 0.58)

= 1 - 0.719

= 0.281

greater than .65 is : 0.281

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