Respuesta :
Answer:
4.18% probability that less than 80% of this sample will devote work time to follow games
Step-by-step explanation:
Normal probability distribution
When the distribution is normal, we use the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
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 [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]
In this question, we have that:
[tex]p = 0.86, n = 100[/tex]
So
[tex]\mu = 0.86, s = \sqrt{\frac{0.86*0.14}{100}} = 0.0347[/tex]
What is the probability that less than 80% of this sample will devote work time to follow games?
This is the pvalue of Z when X = 0.8. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.8 - 0.86}{0.0347}[/tex]
[tex]Z = -1.73[/tex]
[tex]Z = -1.73[/tex] has a pvalue of 0.0418
4.18% probability that less than 80% of this sample will devote work time to follow games
