Rachel received the following scores on the midterm and final examinations in his introductory psychology course. The mean and the standard deviation for each test are given below. Assume that the...


 Rachel received the following scores on the midterm and final examinations in his introductory psychology course. The mean and the standard deviation for each test are given below. Assume that the normal distribution provides a good model for the frequency distribution.




























Section





X





M





s



Midterm



40



50



8



Final



90



100



12






What percentage of students scored better than Rachel on the midterm?


*is there any way to not use excel? I would like a more simpler broken down version. I wasn't sure how to breakdown the equation from the previous answer in the photo attached*


10:34 C<br>= bartleby<br>E Q&A<br>Math / Statistics / Q&A Library / Rachel received the follow<br>Rachel received the following scores on t.<br>The percentage of students scored better than<br>Rachel on the midterm is obtained below as<br>follows:<br>The required probability is,<br>P(X >40) =1- P(X < 40)<br>From EXCEL<br>=1- P(Z <-1.25)|<br>NORMSDIST (–1.25)]<br>=1-0.1057<br>= 0.8943<br>The percentage of students scored better than<br>Rachel on the midterm is 89.43%.(=0.8943)<br>AA<br>A bartleby.com<br>

Extracted text: 10:34 C = bartleby E Q&A Math / Statistics / Q&A Library / Rachel received the follow Rachel received the following scores on t. The percentage of students scored better than Rachel on the midterm is obtained below as follows: The required probability is, P(X >40) =1- P(X < 40)="" from="" excel="1-" p(z=""><-1.25)| normsdist="" (–1.25)]="1-0.1057" =="" 0.8943="" the="" percentage="" of="" students="" scored="" better="" than="" rachel="" on="" the="" midterm="" is="" 89.43%.(="0.8943)" aa="" a="">

Jun 10, 2022
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