3. 150) Math major has a mean GP.A of 298 with a standard deviation or 0.4. Assu

Question

3. 150) Math major has a mean GP.A of 298 with a standard deviation or 0.4. Assume that the GPA distribution is approximately normal (a) What proportion of Math students have GPA more than 3.3 -lab #3.3-218-0.8 .78 o. 1 (b) GPA being 3.3 is EXCEED, during an EXCEED meeting, you meet a bright young woman, young lady's GPA i higher tham Mg, you met bright young woman, what is the chance that the a requirement for students to join an honors Society in the field of math called Z 38-218 3.833) Px38) PX33 o,4 9149 -7881 1917 (c) A random sample of 10 Math majors were collected. What is the chance that exactly two students are a member ofEXCEED 31 032 5 199 (d) About w hat proportion of the math pajors has a GPA between 3 and 3.5? 3.S 2-3 (e) A scholarship were established for students with top 2% of GPA what is the minimum GPA one has to have in order to receive the scholarshrip?

Explanation / Answer

a) here we want to find P(X > 3.3) = 1 - P(X < 3.3) ....1)

Z score corresponding to X = 3.3 is 0.8

so from z table P(X < 3.3) = P(Z < 0.8) = 0.7881

Plug this value in equation 1), so we get

P(X > 3.3) = 1 - 0.7881 = 0.2119 this is the final answer of part a).

b) Here we want to find P(X > 3.8 | X > 3.3) = P(X > 3.8, X > 3.3) / P(X > 3.8) = P(X > 3.8)/ P(X > 3.3)

Let's find P(X > 3.8) = 1 - P(X < 3.8) =1- P(Z < 2.05) =1- 0.9798 = 0.0202

From part a) P( X > 3.3) = 0.2119

So that P(X > 3.8 | X > 3.3) = 0.0202/0.2119 = 0.0953

c) n = sample size = 10, X = number of success = 2 and p = 0.2119

We want to find P(X = 2) using binomial distribution

10C2 = 45

p2 = 0.21192 = 0.0449

(1-p)^8 = (1 - 0.2119)^8 = = 0.1488

So P(X = 2) = 45*0.0449*0.1488 = 0.3007

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