Suppose the results for stat 201 in 2014 are normally distributed with mean 160 and
standard deviation 18 . Find the probability that a random sample of size 36 gives a
sample mean value in the following cases
Question : between 120 and 220
p(120 < z < 220)
p(120-160/3 < z < 220-160/ 3)
p(-13.33 < z < 20)
got stuck...
try
http://davidmlane.com/hyperstat/z_table.html
To find the probability that a random sample of size 36 gives a sample mean value between 120 and 220, you need to use the standard normal distribution and convert the given values into z-scores.
The formula to calculate the z-score is:
z = (x - μ) / (σ / sqrt(n))
where x is the given value, μ is the mean of the population, σ is the standard deviation of the population, and n is the sample size.
For the given problem:
μ = 160 (mean of the population)
σ = 18 (standard deviation of the population)
n = 36 (sample size)
First, let's calculate the z-scores for the values 120 and 220:
For 120:
z1 = (120 - 160) / (18 / sqrt(36))
= -40 / (18 / 6)
= -40 / 3
= -13.33
For 220:
z2 = (220 - 160) / (18 / sqrt(36))
= 60 / (18 / 6)
= 60 / 3
= 20
Now we need to find the probability that the z-score falls between -13.33 and 20. We can look up the probabilities corresponding to these z-scores from a standard normal distribution table or use a statistical software.
p(-13.33 < z < 20) is the probability that the z-score falls between -13.33 and 20.
To find this probability, you can subtract the cumulative probability for -13.33 from the cumulative probability for 20.
P(-13.33 < z < 20) = P(z < 20) - P(z < -13.33)
Using a standard normal distribution table or software, you can find the probabilities corresponding to the z-scores (P(z < 20) and P(z < -13.33)) and subtract them to get the final probability.
Remember that if you are using a standard normal distribution table, you need to convert the z-scores into their corresponding cumulative probabilities.