Q: Mean of taking a test is 22.3 minutes with a standard deviation of 2.8 minutes what probability will fall between 18 and 23.1 minutes?

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if standard deviation is 4 minutes 95% probability is about 2 standard deviations (actually 1.96) so you would need to allow 30 + 8 = 38 minutes

The probability of a phone being answered in 2 minutes, given that the average time is 3 minutes, is not specified in the information given. More details or specific probabilities are needed to determine the answer.

Mean = 100 Standard Deviation = 25 (120 -100) / 25 = 0.80 th z value of .80 from the table = .2881 P(100-120) = 28.81% What percentage of calls are completed in less than 60 minutes ? (60 - 100) / 25 = - 1.6 from the table the z value is .4452 (one side), then : .5000 - .4452 = .0548 P(60) = 5.48%

There is no information on how much time can be spent on a call.

Assuming a normal distribution: For the fastest 5% we need to find the z value which gives 100% - 5% = 95% of the area under the normal curve (from -∞). Using single tailed tables, we need the z value which gives 95% - 50% = 0.45 (above the mean); this is found to be z ≈ 1.645 z = (value - mean)/standard deviation → value = mean + z × standard deviation ≈ 5 min 17 sec + 1.645 × 12 sec ≈ 5 min 17 sec + 20 s = 5 min 37 sec

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if standard deviation is 4 minutes 95% probability is about 2 standard deviations (actually 1.96) so you would need to allow 30 + 8 = 38 minutes

Time is a continuous variable and so the probability that an event takes any particular value is always 0.

The probability of a phone being answered in 2 minutes, given that the average time is 3 minutes, is not specified in the information given. More details or specific probabilities are needed to determine the answer.

You need to ask 1 question at a time, rewording it into 200 characters or less - it helps you to actually think about what is required instead of getting others to do your work for you. The probability is the area under the normal curve between -∞ and the z-value corresponding to 11:00 for a mean of 9:45 and a standard deviation of 65 minutes. When using single tailed tables, they give the area between the mean and the z-value, ranging from 0 to 0.5. So to find the area between -∞ and the z-value it is the value form the single-tailed table of the z value plus 0.5 For working with time, it will be easier to work in minutes. 9:45 to 11:00 is 1 hr 15 minutes = 75 minutes = x - µ z = (x - µ)/standard deviation = 75/65 ≈ 1.154 → area using single-tailed tables is approx 0.3758 → probability ≈ 0.5 + 0.3757 = 0.8757 or approx 87.6%

Mean = 100 Standard Deviation = 25 (120 -100) / 25 = 0.80 th z value of .80 from the table = .2881 P(100-120) = 28.81% What percentage of calls are completed in less than 60 minutes ? (60 - 100) / 25 = - 1.6 from the table the z value is .4452 (one side), then : .5000 - .4452 = .0548 P(60) = 5.48%

There is no information on how much time can be spent on a call.

The expected number is 500.

The probability is (51.5-51.25)/(52-50) = 0.25/2 = 0.125

It has been determined that the wait-time for computer support is normally distributed with a mean of 30 minutes and a standard deviation of 5 min. If 100 people call, how many would you expect to wait more than 30 min? (use logic here)

Between 1 hour and 1 hour, 20 minutes.

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Assuming a normal distribution: For the fastest 5% we need to find the z value which gives 100% - 5% = 95% of the area under the normal curve (from -∞). Using single tailed tables, we need the z value which gives 95% - 50% = 0.45 (above the mean); this is found to be z ≈ 1.645 z = (value - mean)/standard deviation → value = mean + z × standard deviation ≈ 5 min 17 sec + 1.645 × 12 sec ≈ 5 min 17 sec + 20 s = 5 min 37 sec