1) The amount of time that customers wait in line during peak hoursat one bank i
ID: 2914930 • Letter: 1
Question
1) The amount of time that customers wait in line during peak hoursat one bank is normally distributed with a mean of 14 minutes and astandard deviation of 4 minutes. The percentage of time that thewaiting time lies between 17 and 19 minutes is equal to the areaunder the standard normal curve between ____ and _____. Fill in the blanks by standardizing the normally distributedvariable.2) Which is larger the area under the standard normal curvebetween -1 and 1 or the area under the standard normal curvebetween 0 and 2. Explain your reasoning.
3) Use a table of areas to find the specified area under thestandard normal curve. The area that lies to the right of0.59
2) Which is larger the area under the standard normal curvebetween -1 and 1 or the area under the standard normal curvebetween 0 and 2. Explain your reasoning.
3) Use a table of areas to find the specified area under thestandard normal curve. The area that lies to the right of0.59
Explanation / Answer
mean, = 14standard deviation, = 4 percentage of time that the waiting time lies between 17 and19 minutes is equal to the area under the standardnormal curve between ____ and _____. P(X<17) P(Z<(X-)/) = P(Z<(17-14)/4) =P(Z<0.75) =0.7734 P(X<19) P(Z<(X-)/) = P(Z<(19-14)/4) =P(Z<1.25) = 0.8944 HENCE, percentage of time that the waiting time lies between17 and 19 minutes is equal to the area under thestandard normal curve between 0.7734 and 0.8944. P(Z<(X-)/) = P(Z<(17-14)/4) =P(Z<0.75) =0.7734 P(X<19) P(Z<(X-)/) = P(Z<(19-14)/4) =P(Z<1.25) = 0.8944 HENCE, percentage of time that the waiting time lies between17 and 19 minutes is equal to the area under thestandard normal curve between 0.7734 and 0.8944. P(X<19) P(Z<(X-)/) = P(Z<(19-14)/4) =P(Z<1.25) = 0.8944 HENCE, percentage of time that the waiting time lies between17 and 19 minutes is equal to the area under thestandard normal curve between 0.7734 and 0.8944. P(Z<(X-)/) = P(Z<(19-14)/4) =P(Z<1.25) = 0.8944 HENCE, percentage of time that the waiting time lies between17 and 19 minutes is equal to the area under thestandard normal curve between 0.7734 and 0.8944.
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