Express Care Service has found that the delivery time for packages is normally distributed, with mean...
The personnel office at a large electronics firm regularly schedules job interviews and maintains records of the interviews. From the past records, they have found that the length of a first interview is normally distributed, with mean μ = 36 minutes and standard deviation σ = 9 minutes. (Round your answers to four decimal places.) (a) What is the probability that a first interview will last 40 minutes or longer? (b) Fifteen first interviews are usually scheduled per day. What...
TThe personnel office at a large electronics firm regularly schedules job interviews and maintains records of the interviews. From the past records, they have found that the length of a first interview is normally distributed, with mean μ = 32 minutes and standard deviation σ = 7 minutes. (Round your answers to four decimal places.) (a) What is the probability that a first interview will last 40 minutes or longer? (b) Five first interviews are usually scheduled per day. What...
the personnel office at a large electronics firm regularly schedules job interviews and maintains records of the interviews. from the last records, they have found the length of the first interview is normally distributed, with mean 34 minutes and standard deviation of 9 minutes A. What is the probability that the first interview will last 40 minutes or longer? B. Seven first interviews are usually scheduled per day. What is the probability that the average length of time for the...
The personnel office at a large electronics firm regularly schedules job interviews and maintains records of the interviews. From the past records, they have found that the length of a first interview is normally distributed, with mean μ = 36 minutes and standard deviation σ = 7 minutes. (Round your answers to four decimal places.)
The time spent waiting in the line is approximately normally distributed. The mean waiting time is 6 minutes and the standard deviation of the waiting time is 2 minutes. Find the probability that a person will wait for more than 8 minutes. Round your answer the four decimal places.
The time spent waiting in the line is approximately normally distributed. The mean waiting time is 5 minutes and the standard deviation of the waiting time is 3 minutes. Find the probability that a person will wait for less than 7 minutes. Round your answer to four decimal places.
The time spent waiting in the line is approximately normally distributed. The mean waiting time is 55 minutes and the standard deviation of the waiting time is 22 minutes. Find the probability that a person will wait for more than 33 minutes. Round your answer to four decimal places.
The time spent waiting in the line is approximately normally distributed. The mean waiting time is 6 minutes and the standard deviation of the waiting time is 2 minutes. Find the probability that a person will wait for more than 9 minutes. Round your answer to four decimal places.
The time spent waiting in the line is approximately normally distributed. The mean waiting time is 5 minutes and the standard deviation of the waiting time is 1 minute. Find the probability that a person will wait for more than 3 minutes. Round your answer to four decimal places.
The time spent waiting in the line is approximately normally distributed. The mean waiting time is 5 minutes and the standard deviation of the waiting time is 2 minutes. Find the probability that a person will wait for more than 1 minute. Round your answer to four decimal places.