Determining sampling distribution of proportion

Reference no: EM1392580

According to a well-known newspaper journal, 58.3% of the 222,900 households in particular county get the Sunday edition of local newspaper. Assume that random samples of 200 households are selected.

In what proportion of the samples will between 55% and 60% of the households get the Sunday edition of the newspaper?

P(0.55 < Ps <0.60) = P(-0.95 < Z < 0.49) = 0.5168

0.49 in the Cumulative Standardized Normal Distribution Table (= 0.6879)

What are the specific steps involved in converting P(0.55 < Ps <0.60) to P(-0.95 < Z < 0.49)?

Reference no: EM1392580

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