4. Given a binomial random variable with n = 100 and p = 0.5, estimate the Pr[X

Question

4. Given a binomial random variable with n = 100 and p = 0.5, estimate the Pr[X 55]. 6. The scores on a finite mathematics test were normally distributed with a mean of 70 and a standard deviation of 15.
[a] The minimum score a student could obtain in order to receive an A is 90. What is the probability a randomly selected student receives an A?
[b] The minimum score a student could obtain in order to pass the test is 55. What is the probability a randomly selected student failed the test?

7. A mathematics department wants to attract the top ten percent of high school students based on the analytic portion of the SAT exam (SAT Math). They plan to send a letter containing academic scholarship information to every high school student in the state who received an SAT Math score in the top ten percent. If the mean is 500, the standard deviation is 100, and the SAT Math scores are rounded to the nearest 10 points, what is the lowest SAT Math score that would be used for sending these letters?

11. A teacher says that the top 10% of the class received an A on the last test. The scores were normally distributed with mean 65 and standard deviation 20. Find the minimum score required to get an A.

15. In major league baseball, the mean batting average is 0.245 with a standard deviation of .01. If there are 300 players in the league, what is the expected number of players with an average of 0.4 or higher? [The last player in the major leagues with a batting average above 0.4 was Ted Williams of the Boston Red Soxs in 1941, with a batting average of 0.406.]

Explanation / Answer

4)mean =np=100*0.5=50

std deviation =(np(1-p))1/2 =5

hence P(X>55)=1-P(Z<(54.5-50)/5)=1-P(Z<0.9)=1-0.8159=0.1841

6)a)P(X>90)=1-P(Z<(90-70)/15)=1-P(Z<1.333)=1-0.9088=0.0912

b)P(X<55)=P(Z<1)=0.8413

7)here for heighest 10%, z=1.28155

hence score =mean +z*std deviation =628~630

11) minimum score =65+20*1.28155~91

15)P(X>0.4)=1-P(Z<(0.4-0.245)/0.01)=1-P(Z<15.5)=1-1=0

hence expected numbre of players =0

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