Suppose that the weight of male babies less than 2 months old is normally distributed with mean 11.5 pounds and standard deviation 2.7 pounds. A sample of 36 babies is selected. What is the probability that the average weight of the samlpe is less than 11.29 pounds?
Use z-scores and a z-table for this problem. Because you are given a sample size, you will need to include the sample size in the calculation:
z = (x - mean)/(sd/√n)
Your data:
z = (11.29 - 11.5)/(2.7/√36)
Finish the calculation. Use the z-table to find the probability (remember that the problem is asking "less than 11.29" when looking at the table).
I hope this will help get you started.
To find the probability that the average weight of the sample is less than 11.29 pounds, we need to use the Central Limit Theorem and the characteristics of a normal distribution.
The Central Limit Theorem states that the distribution of sample means will be approximately normally distributed, regardless of the shape of the population's distribution, as long as the sample size is sufficiently large (typically n > 30).
In this case, the mean weight of male babies less than 2 months old is normally distributed with a mean of 11.5 pounds and a standard deviation of 2.7 pounds. Since the sample size is 36 (which is larger than 30), we can assume that the sample mean weight will also be normally distributed.
Now, we need to standardize the sample mean weight to calculate the probability using the standard normal distribution.
The formula for standardizing a value x from a normal distribution with mean μ and standard deviation σ is:
Z = (x - μ) / σ
In this case, we want to calculate the probability that the sample mean weight (x̄) is less than 11.29 pounds, so we need to standardize 11.29 using the given population mean (μ) of 11.5 and standard deviation (σ) of 2.7.
Z = (11.29 - 11.5) / (2.7 / √36)
Simplifying, we get:
Z = (-0.21) / (2.7 / 6)
Z = (-0.21) / 0.45
Z = -0.47
Now that we have the standardized value (Z-score), we can use the standard normal distribution table or a calculator to find the probability associated with this Z-score.
Using the standard normal distribution table, the probability of a Z-score less than -0.47 is approximately 0.3212.
Therefore, the probability that the average weight of the sample is less than 11.29 pounds is approximately 0.3212, or 32.12%.
To solve this problem, we need to use the Central Limit Theorem, which states that the sampling distribution of the sample means approaches a normal distribution, regardless of the shape of the population distribution.
The mean of the sampling distribution of sample means is equal to the population mean, while the standard deviation of the sampling distribution (also known as the standard error) is equal to the population standard deviation divided by the square root of the sample size.
To calculate the probability that the average weight of the sample is less than 11.29 pounds, we need to find the z-score corresponding to this value and then use a standard normal distribution table (z-table) to find the probability.
The z-score can be calculated using the formula:
z = (x - μ) / (σ / √n)
where:
x = average weight of the sample (11.29 pounds)
μ = population mean (11.5 pounds)
σ = population standard deviation (2.7 pounds)
n = sample size (36)
Let's calculate the z-score:
z = (11.29 - 11.5) / (2.7 / sqrt(36))
z = -0.21 / (2.7 / 6)
z = -0.21 / 0.45
z ≈ -0.467
Now, we need to find the area under the standard normal curve to the left of this z-score.
Looking up the z-score -0.467 in the z-table, we find that the corresponding cumulative probability is 0.3207.
Therefore, the probability that the average weight of the sample is less than 11.29 pounds is approximately 0.3207 or 32.07%.