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Converting P(0.15 ≤ p ≤ 0.19) to the standard normal random variable z for a sample...

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Statistics

Converting P(0.15 ≤ p ≤ 0.19) to the standard normal randomvariable z for a sample of 500 households gives P(−1.19 ≤ z ≤1.19). This is the probability that a sample of 500 households willprovide a sample proportion within 0.02 of the populationproportion, 0.17, of households that spend more than $100 per weekon groceries. Use a table to compute P(−1.19 ≤ z ≤ 1.19), roundingthe result to four decimal places. P(−1.19 ≤ z ≤ 1.19) = P(z ≤1.19) − P(z ≤ −1.19) = 0.8830 − ???= ???

Answer & Explanation Solved by verified expert
3.7 Ratings (321 Votes)

P(-1.19 < Z < 1.19)

= P(Z < 1.19) - P(Z < -1.19)

= 0.8830 - 0.1170

= 0.7660

                                         

                                     

                                                      

                       

                                 

                                       

                                  
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