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In 2018 the scores of students on the May SAT had a normal distribution with mean u = 1450 and a standard deviation of o = 120.a. What is the probability that a single student randomly chosen from all those taking the test scores 1500 or higher?b. If a sample of 50 students is taken from the population, what is the probability that the sample mean score of these students is 1470 or higher?

User Arsalan
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From the question,


\begin{gathered} \mu\text{ = 1450, } \\ \sigma\text{ = 120} \end{gathered}

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


\begin{gathered} P(x<\text{ z) such that } \\ z\text{ = }\frac{x\text{ -}\mu}{\sigma} \end{gathered}

From the question, x = 1500.

Therefore


\begin{gathered} z\text{ =}\frac{1500\text{ - 1450}}{120} \\ z\text{ = }(50)/(120) \\ z\text{ = 0.417} \end{gathered}

applying z - test


\begin{gathered} P(x<strong>Thus, the probability that a single student is randomly chosen from all those taking the test scores 1500 or higher is approximately 34%</strong><p></p><p></p><p>b. From the question</p>[tex]\begin{gathered} n\text{ = 50, }^{}\text{ }\mu\text{ = 1450} \\ \bar{x}\text{ = 1470},\text{ }\sigma\text{ = 120} \end{gathered}

we will be using


\begin{gathered} z\text{ = }\frac{\bar{x}\text{ - }\mu}{\frac{\sigma}{\sqrt[]{n}}} \\ \end{gathered}

inserting values


\begin{gathered} z\text{ = }\frac{1470\text{ - 1450}}{\frac{120}{\sqrt[]{50}}} \\ z\text{ = }20\text{ }*\frac{\sqrt[]{50}}{120} \\ z\text{ = }\frac{\sqrt[]{50}}{6} \\ z\text{ = 1.18} \end{gathered}

Applying z-test

[tex]\begin{gathered} P(x

Hence,

The probability that the sample mean score of these students is 1470 or higher is approximately 12%

User Lindyblackburn
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