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s2008m [1.1K]
3 years ago
9

Final exam scores are normally distributed with a mean of 74 and a standard deviation of 6. Approximately, what percentage of fi

nal exam scores would be between 68 and 86?
Mathematics
1 answer:
notka56 [123]3 years ago
3 0

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.

z=\frac{x-\mu}{\sigma}

z=\frac{68-74}{6}

z=\frac{-6}{6}=-1

Now let us find z-score for 86.

z=\frac{86-74}{6}    

z=\frac{12}{6}=2        

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,    

P(-1

P(2>Z)=.97725  

We will use formula P(a 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,

P(-1

P(-1

Upon converting 0.81859 to percentage we will get

0.81859*100=81.859\approx 81.86

Therefore, 81.86% of final exam score will be between 68 and 86.  


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