I am not sure where to go with this 2 part question. Ifsomeone would not mind sh

Question

I am not sure where to go with this 2 part question. Ifsomeone would not mind showing me how to work through thisproblem.... Thanks Q: The time required to assemble a product part are normallydistributed with a mean of 47.5 minutes and a standard deviationequal to 8.5 minutes. 1- What percent of the assembly workers require: more than onehour? Ans. to choose from: 7.08%, 6.03%, 1.97%& 2.59% 2- What percent of the assembly workers require: less than onehalf hour? Ans. to choose from: Same asabove. I keep pluging these numbers into diffrent formulas and cannot come up with the correct solution. Please show me how tocome to the correct solution. Thank You I am not sure where to go with this 2 part question. Ifsomeone would not mind showing me how to work through thisproblem.... Thanks Q: The time required to assemble a product part are normallydistributed with a mean of 47.5 minutes and a standard deviationequal to 8.5 minutes. 1- What percent of the assembly workers require: more than onehour? Ans. to choose from: 7.08%, 6.03%, 1.97%& 2.59% 2- What percent of the assembly workers require: less than onehalf hour? Ans. to choose from: Same asabove. I keep pluging these numbers into diffrent formulas and cannot come up with the correct solution. Please show me how tocome to the correct solution. Thank You

Explanation / Answer

More than 1 hr = 1 - P(less than 1 hr) 1 hr = 60 min --> z = (X - ) / = (60 - 47.5) / 8.5= 1.47 Using a z table, we see this has an area to the left of p =.9292. However, as stated earlier, since we want the area tothe RIGHT, we subtract this from 1 and get 1 = .9292 = .0708 or7.08% 1 half hour = 30 min z(30) = (30 - 47.5) / 8.5 = -2.06 Using a z table, we see this has an area to the left of .0197 or1.97%.

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