The length of country and western songs is normally distributed and has a mean of 200 seconds and a standard deviation of 30 seconds. Find the probability that a random selection of 9 songs will have mean length of 186.30 seconds or less. Assume the distribution of the lengths of the songs is normal.
z = (186.3-200)/(30/sqrt(9))
z = -1.37, you can use your table
The average age of doctors in a certain hospital is 50.0 years old. Suppose the distribution of ages is normal and has a standard deviation of 4.0 years. If 16 doctors are chosen at random for a committee, find the probability that the average age of those doctors is less than 50.3 years. Assume that the variable is normally distributed.
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To find the probability that a random selection of 9 songs will have a mean length of 186.30 seconds or less, we first need to calculate the standard deviation of the sample mean.
The standard deviation of the sample mean, also known as the standard error, is given by the formula:
Standard Error = Standard Deviation / sqrt(n)
where n is the sample size.
In this case, the given standard deviation is 30 seconds, and the sample size is 9 songs. Plugging these values into the formula:
Standard Error = 30 / sqrt(9) = 30 / 3 = 10
Next, we need to calculate the z-score, which measures the number of standard deviations a given value is from the mean. The z-score is calculated using the formula:
z = (x - mean) / standard error
where x is the given value, mean is the mean of the population, and standard error is the standard error of the sample mean.
In this case, the given value is 186.30 seconds, the mean of the population is 200 seconds, and the standard error is 10 seconds. Plugging these values into the formula:
z = (186.30 - 200) / 10 = -13.7 / 10 = -1.37
Now, we need to find the probability corresponding to this z-score using a standard normal distribution table or a calculator. The probability can be interpreted as the area under the normal distribution curve to the left of the given z-score.
Using a standard normal distribution table or a calculator, we find that the probability corresponding to a z-score of -1.37 is approximately 0.0841.
Therefore, the probability that a random selection of 9 songs will have a mean length of 186.30 seconds or less is approximately 0.0841, or 8.41%.