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Answer:
0.997 = 99.7% probability that the resulting sample proportion to be between 0.066 and 0.294
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
When the distribution is normal, we use the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean and standard deviation
18% of residents in Minneapolis, Minnesota, are African Americans. Suppose a local court will randomly sample 100 state residents and will then observe the proportion in the sample who are African American.
This means that
So, by the Central Limit Theorem:
How likely is the resulting sample proportion to be between 0.066 and 0.294?
This is the pvalue of Z when X = 0.294 subtracted by the pvalue of Z when X = 0.066. So
X = 0.294
By the Central Limit Theorem
has a pvalue of 0.9985
X = 0.066
has a pvalue of 0.0015
0.9985 - 0.0015 = 0.997
0.997 = 99.7% probability that the resulting sample proportion to be between 0.066 and 0.294
Answer:
The probability that X will be more than 0.5 away fro the population
mean is 0.7812 ⇒ answer a
Step-by-step explanation:
* Let us explain how to solve the problem
- The sample size n is 25
- The mean of the population μ is 41
- The standard deviation σ is 9
- We need to find the approximate probability that X will be more than
0.5 away from the population mean
∵ z-score = (X - μ)/(σ/√n)
∵ X - μ = 0.5
- That means X > 41 + 0.5, OR X < 41 - 0.5
- Then we need to find P(X > 41.5) and P(X < 40.5)
∵ z = = 0.2778
∵ z = = -0.2778
- Let us use the normal distribution table to find the corresponding
area to z-score
∵ P(z < 0.2778) = 0.6116
∵ P(z > -0.2778) = 0.3928
- P(40.5 < X < 41.5) = P( -0.2778 < z < 0.2778)
∴ The probability of X to be with in 0.5 = 0.6116 - 0.3928
∴ The probability of X to be with in 0.5 = 0.2188
∵ The probability of X to be more than 0.5 away from the population
mean = 1 - 0.2188
∴ P(X > 41.5) OR P(X < 40.5) = 0.7812
* The probability that X will be more than 0.5 away fro the population
mean is 0.7812