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 sample is less than 11.07 pounds?
Write only a number as your answer. Round to 4 decimal places (for example 0.0048). Do not write as a percentage.
**Please explain how to do it****
Z = (sample-mean)/SEm
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
To solve this problem, we will use the Central Limit Theorem, which states that if a sample size is large enough, the distribution of sample means will be approximately normally distributed, regardless of the shape of the population distribution.
Step 1: Calculate the standard deviation of the sample mean (also known as the standard error). The formula for the standard error is:
Standard Error = Standard Deviation / √(Sample Size)
In this case, the standard deviation of the population is given as 2.7 pounds, and the sample size is 36.
Standard Error = 2.7 / √(36) = 2.7 / 6 = 0.45
Step 2: Convert the given value of the sample mean to the z-score using the formula:
z = (x - μ) / σ
where x is the given value (11.07 pounds), μ is the population mean (11.5 pounds), and σ is the standard error (0.45).
z = (11.07 - 11.5) / 0.45 = -0.43 / 0.45 ≈ -0.96
Step 3: Use the z-table or a statistical calculator to find the probability corresponding to the z-score obtained in step 2.
Looking up the z-score of -0.96 in the z-table, we find the corresponding probability to be approximately 0.1675.
Therefore, the probability that the average weight of the sample is less than 11.07 pounds is approximately 0.1675.
Remember to round the answer to four decimal places as specified in the question, so the final answer is 0.1675.