A telephone company receives 3 calls every minute on an average.
If an operator has a watch that is not very accurate and which has a normal distribution with mean of 1 minute and standard deviation of 5 seconds, what is the probability that he measures 5 calls in a minute?
To find the probability that the operator measures 5 calls in a minute, we can use the concept of the normal distribution.
Step 1: Convert the problem to a standard normal distribution
We know that the operator receives 3 calls every minute on average. So, in 1 minute, the operator is expected to receive 3 calls. We need to calculate the probability of receiving 5 calls, which is 2 more than the average.
To convert this problem to a standard normal distribution, we can calculate the z-score. The formula for calculating the z-score is:
z = (x - μ) / σ
where x is the observed value, μ is the mean, and σ is the standard deviation.
In this case, the observed value is 5, the mean is 3, and the standard deviation is given as 5 seconds, which is equal to 5/60 minutes. So the z-score is:
z = (5 - 3) / (5/60) = 12
Step 2: Find the probability using the standard normal distribution table
Now that we have the z-score, we can use a standard normal distribution table to find the corresponding probability.
The probability associated with the z-score of 12 can be found by looking up the value in the table or using a calculator that has this functionality. However, standard normal distribution tables typically only provide z-scores up to a certain value, often 3 or 4. Since the z-score of 12 falls well beyond this range, we can assume that the probability is extremely close to 1 (or 100%).
Therefore, the probability that the operator measures 5 calls in a minute is approximately 1 (or 100%).
Note: It's important to mention that the operator's watch being inaccurate does not affect the probability calculation. The normal distribution is based on the assumption that the average and standard deviation are known and constant. The inaccuracy of the watch only comes into play when estimating the mean and standard deviation of the distribution itself, but not when calculating the probability using those estimated values.