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Assume the average test score from the last year's class was 85 points with a standard deviation of 9. By sketching a normal curve, which value below is the best estimate for the proportion of students earned a 70 points or higher?

Assume the average test score from the last year's class was 85 points with a standard-example-1

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The given information is:

- The average test score is 85 points

- The standard deviation is 9

- It is a normal distribution

We need to find the best estimate for the proportion of students earned a 70 points or higher.

First, we need to find the z-score for 70, by applying the following formula:


z=(x-\mu)/(\sigma),\text{ where }\mu=average,\sigma=standard\text{ }deviation,x=value

Then, as x=70, we replace the given values and find z:


z=(70-85)/(9)=(-15)/(9)=-1.67

Then, we can find the estimate as follows:


P(X\ge70)=P(z\ge-1.67)=1-P(z<-1.67)

Now, in a standard normal table, we search the cumulative probability of z<-1.67, and it is 0.0475:

Thus, replace this value to find P(X>=70):


P(X\ge70)=1-0.0475=0.9525

The probability is 0.9525, if we multiply it by 100%, we obtain: 95.25%

The best estimate is 95%.

Assume the average test score from the last year's class was 85 points with a standard-example-1
User James Leonard
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