The number of boxes of girl scout cookies sold by each of the girl scout in a city has a distribution which is approximately normal with a mean=75 boxes and a standard deviation=30 boxes....
1.) find the probability that scout chosen at random sold between 60 and 90 boxes.
2.) Find the probability that the sample mea number of boxes of cookies sold by a random sample of 36 scouts is between 60 and 90 boxes.
My answers:
1.) .3830
2.) .9974
Just need to know if I am getting the right answers and if not then need the correct way to calculate....Im getting confused.
Both answers are right! Good work!
78% of U.S. homes have a telephone answering device. In a random sample of 268 homes, what is the probability that fewer than 53 do not have a telephone anwering device?
To calculate the probabilities, you need to use the standard normal distribution. The standard normal distribution has a mean of 0 and a standard deviation of 1.
To solve the problems, you will need to standardize the values using the z-score formula:
z = (X - μ) / σ
Where:
X = Value you want to standardize
μ = Mean of the distribution
σ = Standard deviation of the distribution
Once you have the standardized values, you can use a standard normal distribution table or a statistical software to find the probabilities associated with those values.
Let's solve the problems step by step:
1. Probability that a randomly chosen scout sold between 60 and 90 boxes:
To find this probability, we need to standardize the values using the given mean and standard deviation.
Standardizing X = 60:
z1 = (60 - 75) / 30 = -0.5
Standardizing X = 90:
z2 = (90 - 75) / 30 = 0.5
Now, we need to find the probability between these two z-scores. You can look up the values in a standard normal distribution table or use a statistical software.
Using the standard normal distribution table, the probability between -0.5 and 0.5 is approximately 0.3830.
Therefore, your answer is correct.
2. Probability of the sample mean:
To solve this problem, we need to use the Central Limit Theorem, which states that the distribution of the sample means approaches a normal distribution with mean μ and standard deviation σ / sqrt(n), where n is the sample size.
Given that n = 36 and the mean and standard deviation are the same as before (75 and 30, respectively), we can calculate the standard deviation of the sample mean:
Standard deviation (sample mean) = σ / sqrt(n) = 30 / sqrt(36) = 5
Now, we repeat the process of standardizing the values:
Standardizing X = 60:
z1 = (60 - 75) / 5 = -3
Standardizing X = 90:
z2 = (90 - 75) / 5 = 3
Using the standard normal distribution table, the probability between -3 and 3 is approximately 0.9974.
Therefore, your answer is correct for the sample mean as well.
In conclusion, your answers are correct. You used the correct approach to calculate these probabilities.