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X 7.3.46-T Question Help If a random sample of 100 items is taken from a population in which the proportion of items having a
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solution:

the given information as follows:

sample size = n = 100

sample proportion or probability of success = p = 0.35

since np = 100*0.35 = 35 and n(1-p) = 100*(1-0.35) = 65 are greater than 10 so we can use the normal distribution

so mean = \mu_{\bar p} = p = 0.35

SD = \sigma_{\bar p}= \sqrt{\frac{p(1-p)}{{n}}}=\sqrt{\frac{0.35(1-0.35)}{100}}=0.0477

we have to find the probability of success is less than or equal to 0.42 = P(\bar p < 0.42)

so corresponding z score

z=\frac{\bar p-\mu_{\bar p}}{\sigma_{\bar p}}=\frac{0.42-0.35}{0.0477}=1.47

P(\bar p < 0.42) = value of z to the left of 1.47 from the z table = 0.9292

so, the probability will be 0.9292

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