Answer:
2.28% of players are under 86kg
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be 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.
The weights of the players were distributed normally with a mean of 98kg and a standard deviation of 6kg.
This means that [tex]\mu = 98, \sigma = 6[/tex]
What percentage of players are under 86kg?
The proportion is the pvalue of Z when X = 86. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{86 - 98}{6}[/tex]
[tex]Z = -2[/tex]
[tex]Z = -2[/tex] has a pvalue of 0.0228
0.0228*100% = 2.28%
2.28% of players are under 86kg
