Question

Mr Emmer gave a test in his Chemistry class. The scores were normally distributed with a mean of 82 and a standard deviation of 4 What percent of students would you expect to score between 74 and 787 (Remember to refer t

normal distribution located on page 8 of the lesson)
12.5%
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Answer 1

Hi,

Explanation:

To find the percentage of students who would be expected to score between 74 and 78 on Mr. Emmer's Chemistry test, you can use the z-score and the standard normal distribution (also known as the Z-distribution).

First, you need to find the z-scores for both 74 and 78 using the given mean (μ=82) and standard deviation (σ=4):

`Z=(X-\mu)/(\sigma) =(X-82)/(4) \\if\ X=90\ then\ Z=(90-82)/(4)=2\\if\ X=86\ then\ Z=(86-82)/(4)=1\\p(X\leq 78) -p(X \leq 74)\\=p(82+(82-74) ) -p(82+(82-78))\\=p(X\leq 90)-p(X\leq 86)\\=p(Z\leq 2)-p(Z\leq 1)\\=0.9772-0.8413\\=0.1359\\=13.59\%`

Now, you want to find the percentage of students who score between -2 and -1 on the standard normal distribution curve. You can do this using a standard normal distribution table or a calculator.

The area between -2 and -1 on the standard normal distribution curve represents the percentage of students in this range.

You can find this using a standard normal distribution table or a calculator. It's typically about 0.1359, which means approximately 13.59% of students would be expected to score between 74 and 78 on Mr. Emmer's Chemistry test.