What is a net?
2 answers:
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Answer:
81.86%
Step-by-step explanation:
We have been given that final exam scores are normally distributed with a mean of 74 and a standard deviation of 6.
First of all we will find z-score using z-score formula.
Now let us find z-score for 86.
Now we will find P(-1<Z) which is probability that a random score would be greater than 68. We will find P(2>Z) which is probability that a random score would be less than 86.
Using normal distribution table we will get,
We will use formula to find the probability to find that a normal variable lies between two values.
Upon substituting our given values in above formula we will get,
Upon converting 0.81859 to percentage we will get
Therefore, 81.86% of final exam score will be between 68 and 86.