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.
Well, you know what they say about doctors getting older - they age like fine wine! 🍷
To find the probability that the average age of those chosen doctors is less than 50.3 years, we can use the central limit theorem. Since the distribution of ages is normal, the distribution of the sample means will also be normal.
The mean of the sample means will still be 50.0 years, and the standard deviation of the sample means (also known as the standard error) can be calculated by dividing the population standard deviation by the square root of the sample size.
So, in this case, the standard error is 4.0 / √16 = 1.0 years.
Now, let's convert the given average age of 50.3 years to a z-score using the formula:
z = (x - μ) / σ
where x is the desired value (50.3 years), μ is the mean (50.0 years), and σ is the standard error (1.0 years).
Plugging in the values, we have:
z = (50.3 - 50.0) / 1.0 = 0.3
Using a standard normal distribution table or calculator, you can find that the probability of obtaining a z-score less than 0.3 is approximately 0.6179.
So, the probability that the average age of the committee is less than 50.3 years is approximately 0.6179 or 61.79%.
Looks like this committee might have a few young guns! 🧓🏻♂️🔫
To find the probability that the average age of the chosen doctors is less than 50.3 years, we need to calculate the z-score and use the standard normal distribution.
Step 1: Calculate the standard error (SE) of the sample mean.
The standard error is the standard deviation divided by the square root of the sample size (n).
SE = standard deviation / √n
SE = 4.0 years / √16
SE = 4.0 years / 4
SE = 1.0 years
Step 2: Calculate the z-score.
The z-score measures how many standard errors a given value is away from the mean.
z-score = (X - μ) / SE
Here, X is the average age we want to calculate the probability for, which is 50.3 years.
μ represents the mean, which is 50.0 years.
SE is the standard error we calculated in Step 1, which is 1.0 years.
z-score = (50.3 - 50.0) / 1.0
z-score = 0.3 / 1.0
z-score = 0.3
Step 3: Look up the z-score in the standard normal distribution table.
The standard normal distribution table provides the probability associated with each z-score. We want to find the probability to the left of the z-score.
Looking up the z-score of 0.3 in the table, we find that the corresponding probability is approximately 0.6179.
Therefore, the probability that the average age of the chosen doctors is less than 50.3 years is approximately 0.6179 or 61.79%.
To solve this problem, we will convert it into a standard normal distribution by using the central limit theorem.
The central limit theorem states that if the sample size is large enough, the distribution of sample means will approach a normal distribution, regardless of the shape of the population distribution.
To find the probability that the average age of those doctors is less than 50.3 years, we need to calculate the z-score and then find the corresponding probability using the standard normal distribution table.
1. Calculate the standard deviation of the sampling distribution (also known as the standard error) using the formula: standard deviation / sqrt(sample size).
Standard deviation = 4.0 years
Sample size = 16
Standard error = 4.0 / sqrt(16) = 1.0 years.
2. Calculate the z-score, which represents the number of standard deviations away from the mean the desired value is. The formula for the z-score is: z = (x - mean) / standard error.
x = 50.3 years
mean = 50.0 years
standard error = 1.0 years.
z = (50.3 - 50.0) / 1.0 = 0.3
3. Use a standard normal distribution table (also known as a z-table) to find the corresponding probability of the given z-score. Look up the z-score (0.3) in the table, which will give you the area under the curve to the left of the z-score.
The probability associated with a z-score of 0.3 is approximately 0.6179.
Therefore, the probability that the average age of those doctors is less than 50.3 years is approximately 0.6179, or 61.79%.