Respuesta :
Answer:
99.89% of students scored below 95 points.
Step-by-step explanation:
Problems of normally distributed samples are solved using 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.
In this problem, we have that:
[tex]\mu = 76.4, \sigma = 6.1[/tex]
What percent of students scored below 95 points?
This is the pvalue of Z when X = 95. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{95 - 76.4}{6.1}[/tex]
[tex]Z = 3.05[/tex]
[tex]Z = 3.05[/tex] has a pvalue of 0.9989.
99.89% of students scored below 95 points.
The percent of students scored below 95 points is 99.89%
Given that,
- Normal distribution with a mean of 76.4 and a standard deviation of 6.1 points.
calculation:
[tex]= (95 - 76.4) \div 6.1[/tex]
= 3.05
It has a p-value of 0.9989.
- The Z-score determined how many standard deviations the measure is from the mean.
- After determining the Z-score, we watch at the z-score table and find the p-value associated with this z-score.
Find out more information about percentage here:
https://brainly.com/question/26080842?referrer=searchResults
