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Using the <em>normal distribution and the central limit theorem</em>, we have that:
a) The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.
b) There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.
c) There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.
<h3>Normal Probability Distribution</h3>In a normal distribution with mean and standard deviation
, the z-score of a measure X is given by:
- It 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, which is the percentile of X.
- By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean
and standard deviation
, as long as
and
.
In this problem:
- 22% of couples meet online, hence p = 0.22.
- A sample of 150 couples is taken, hence n = 150.
Item a:
The mean and the standard error are given by:
The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.
Item b:
The probability is <u>one subtracted by the p-value of Z when X = 0.25</u>, hence:
By the Central Limit Theorem:
Z = 0.89
Z = 0.89 has a p-value of 0.8133.
1 - 0.8133 = 0.1867.
There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.
Item c:
The probability is the <u>p-value of Z when X = 0.2 subtracted by the p-value of Z when X = 0.15</u>, hence:
X = 0.2:
Z = -0.59
Z = -0.59 has a p-value of 0.2776.
X = 0.15:
Z = -2.07
Z = -2.07 has a p-value of 0.0192.
0.2776 - 0.0192 = 0.2584.
There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.
To learn more about the <em>normal distribution and the central limit theorem</em>, you can check brainly.com/question/24663213