A recent survey of 550 residents in large city indicated that 65% of the residents had...

A recent survey of 550 residents in large city indicated that 65% of the residents had health

insurance. Suppose that it is indeed true that 65% of the people have health insurance. The

sampling distribution for the proportion of people with health insurance is approximately normal

with mean 0.65 and standard deviation 0.02.

What is the probability the sample proportion will be greater than 0.68?

Solutions

Expert Solution

Solution

Given that,

p = 0.65

1 - p = 1 -0.65=0.35

n = 550

= p =0.65

= $\sqrt$ [p( 1 - p ) / n] = $\sqrt$ [0.65*(0.35) /550 ] = 0.02

P( > 0.68) = 1 - P( <0.68 )

= 1 - P(( - ) / < (0.68 -0.65) /0.02 )

= 1 - P(z < 1.5)

= 1 - 0.9332 = 0.0668

probability = 0.0668

orchestra answered 2 years ago

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