IDNLearn.com: Where your questions meet expert advice and community support. Our experts provide timely and accurate responses to help you navigate any topic or issue with confidence.
Sagot :
From the question,
[tex]\begin{gathered} \mu\text{ = 1450, } \\ \sigma\text{ = 120} \end{gathered}[/tex]a. We are to find the probability that a single student randomly chosen from all those taking the test scores 1500 or higher?
we will do this using
[tex]\begin{gathered} P(x<\text{ z) such that } \\ z\text{ = }\frac{x\text{ -}\mu}{\sigma} \end{gathered}[/tex]From the question, x = 1500.
Therefore
[tex]\begin{gathered} z\text{ =}\frac{1500\text{ - 1450}}{120} \\ z\text{ = }\frac{50}{120} \\ z\text{ = 0.417} \end{gathered}[/tex]applying z - test
[tex]\begin{gathered} P(xThus, the probability that a single student is randomly chosen from all those taking the test scores 1500 or higher is approximately 34%b. From the question
[tex]\begin{gathered} n\text{ = 50, }^{}\text{ }\mu\text{ = 1450} \\ \bar{x}\text{ = 1470},\text{ }\sigma\text{ = 120} \end{gathered}[/tex]we will be using
[tex]\begin{gathered} z\text{ = }\frac{\bar{x}\text{ - }\mu}{\frac{\sigma}{\sqrt[]{n}}} \\ \end{gathered}[/tex]inserting values
[tex]\begin{gathered} z\text{ = }\frac{1470\text{ - 1450}}{\frac{120}{\sqrt[]{50}}} \\ z\text{ = }20\text{ }\times\frac{\sqrt[]{50}}{120} \\ z\text{ = }\frac{\sqrt[]{50}}{6} \\ z\text{ = 1.18} \end{gathered}[/tex]Applying z-test
[tex]\begin{gathered} P(xHence,
The probability that the sample mean score of these students is 1470 or higher is approximately 12%
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. Find reliable answers at IDNLearn.com. Thanks for stopping by, and come back for more trustworthy solutions.
