A random sample of size 36 is taken from a population with mean 50 and standard deviation 5. Find P( < 51.5).
.9641
To solve this problem, we can use the standard normal distribution and the formula for z-score.
Step 1: Calculate the standard error (SE)
The standard error (SE) is the standard deviation of the sample mean, and it is calculated by dividing the population standard deviation by the square root of the sample size.
SE = σ / √n
= 5 / √36
= 5 / 6
= 0.8333 (rounded to 4 decimal places)
Step 2: Calculate the z-score
The z-score measures the number of standard deviations an individual value is from the mean. In this case, we want to find the z-score for x = 51.5.
z = (x - μ) / SE
= (51.5 - 50) / 0.8333
= 1.5 / 0.8333
≈ 1.8
Step 3: Find the probability using a standard normal distribution table or calculator
Using a standard normal distribution table or calculator, we can find that the probability of z < 1.8 is approximately 0.9641.
Therefore, P(x < 51.5) ≈ 0.9641 or 96.41%.
To find P(X < 51.5) where X is the random variable representing the sample mean, we can use the central limit theorem.
The central limit theorem states that for a large enough sample size, the sample mean follows an approximately normal distribution, regardless of the shape of the population distribution. In this case, we have a random sample of size 36, which is considered large enough for the central limit theorem to apply.
Given that the population mean is 50 and the standard deviation is 5, we can find the standard deviation of the sample mean (also known as the standard error) using the formula:
Standard Error = Population Standard Deviation / Square Root of Sample Size
Standard Error = 5 / √36 = 5 / 6 = 0.8333
Now, we can convert the original problem into a z-score problem by calculating the z-score using the formula:
z = (X - Population Mean) / Standard Error
z = (51.5 - 50) / 0.8333 = 1.80
Finally, we can find the probability P(X < 51.5) by looking up the z-score in the standard normal distribution table or using a calculator. The area to the left of the z-score of 1.80 represents the probability.
Using a standard normal distribution table or a calculator, we find that P(Z < 1.80) is approximately 0.9641.
Therefore, P(X < 51.5) is approximately 0.9641, or 96.41%.
If this is for distribution of means,
Z = (score-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.