Using the normal distribituion, it is found that 2.28% of of scores were less than 40.
Normal Probability Distribution
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- 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.
In this problem, the mean and the standard deviation are given by, respectively:
[tex]\mu = 52, \sigma = 6[/tex].
The proportion of scores that were less than 40 is the p-value of Z when X = 40, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{40 - 52}{6}[/tex]
[tex]Z = -2[/tex]
[tex]Z = -2[/tex] has a p-value of 0.0228.
2.28% of of scores were less than 40.
More can be learned about the normal distribituion at https://brainly.com/question/24663213
