what is the probability that a random sample of 25 pregnancies has a mean gestation period of 258 days or less? standard deviation of 16 days.
To solve this problem, we can use the central limit theorem, which states that the distribution of sample means approximates a normal distribution as the sample size increases.
Given:
Sample size (n) = 25
Mean (μ) = 258 days
Standard deviation (σ) = 16 days
First, calculate the standard error (SE) using the formula:
SE = σ / √n
SE = 16 / √25
SE = 16 / 5
SE = 3.2
Next, calculate the z-score using the formula:
z = (x - μ) / SE
where x is the value we are interested in (258 days).
z = (258 - 258) / 3.2
z = 0
Since the z-score is 0, the probability is equivalent to the area under the curve to the left of 0 in the standard normal distribution. Using a z-table or a statistical software, this probability is found to be approximately 0.5000.
Therefore, the probability that a random sample of 25 pregnancies has a mean gestation period of 258 days or less is 0.5000 or 50%.
To find the probability that a random sample of 25 pregnancies has a mean gestation period of 258 days or less, we need to use the Central Limit Theorem.
The Central Limit Theorem states that for a sufficiently large sample size, the distribution of sample means will be approximately normally distributed regardless of the shape of the population distribution.
In this case, the sample size is 25, which is considered large enough for the Central Limit Theorem to apply.
To calculate the probability, we can standardize the sample mean using the z-score formula:
z = (x - μ) / (σ / sqrt(n))
Where:
- z is the z-score
- x is the mean gestation period we are interested in (258 days)
- μ is the population mean gestation period (unknown in this case)
- σ is the population standard deviation (16 days)
- n is the sample size (25)
Since we don't know the population mean, we need to make an assumption. Let's assume that the population mean gestation period is 280 days.
Using the formula, we can calculate the z-score:
z = (258 - 280) / (16 / sqrt(25))
z = -22 / (16 / 5)
z = -22 / 3.2
z = -6.875
Next, we can use a standard normal distribution table or a statistical software to find the probability associated with the z-score of -6.875.
Using either method, we find that the probability is extremely close to 0, indicating that it is highly unlikely to have a sample mean gestation period of 258 days or less for a random sample of 25 pregnancies, assuming a population mean of 280 days and a population standard deviation of 16 days.