Standard for ola (y + 3)2 = -12(x - 2), iden 1. D n111
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Answer:
0.6214 = 62.14% probability that the sample proportion is between 0.26 and 0.38
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score 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 p-value, 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
37% of the company's orders come from first-time customers.
This means that
A random sample of 225 orders will be used to estimate the proportion of first-time-customers.
This means that
Mean and standard deviation:
What is the probability that the sample proportion is between 0.26 and 0.38?
This is the pvalue of Z when X = 0.38 subtracted by the pvalue of Z when X = 0.26.
X = 0.38
By the Central Limit Theorem
has a pvalue of 0.6217
X = 0.26
has a pvalue of 0.0003
0.6217 - 0.0003 = 0.6214
0.6214 = 62.14% probability that the sample proportion is between 0.26 and 0.38