chapt. 7. please answer asap!!

Canverting P(0.15p0.19) to the standard normal random variable z for a sample of 900 households gives P(1.60z1.60), This is the probability that a sampie of 900 households will provice a sample proportion within 0.02 of the population prooortion, 0.17, of households that spend mare than 5100 . per week on groceries. Use a table to compute P(1.60z1.60), rounding the result to four decimal places. P(1.60z1.60)=P(z1.60)P(z1.60)=0.9452=