a standarized test has a mean of 500 with a standard deviation of70. Assume 3000

Question

a standarized test has a mean of 500 with a standard deviation of70. Assume 3000 students took the test and their scoresapproximated a normal curve. Illustrate this information on anormal curve and answer the following questions assuming 3000students took the test and their scores approximated a normalcurve.
a) About how many students scored between 360 and 640?
b) A score above 640 is considered exceptional on this exam. Abouthow many students made exceptional scores?
c) About how many more students received scores between 570 and 640than between 290 and 360?
d) If a student received a score of 600, at approximately whatpercentile did she score?

Explanation / Answer

mean , = 500
standard deviation , = 70 n = 3000 a) About how many students scored between 360 and 640 P(X<360) = P(Z<(X-)/)=P(Z<(360-500)/70) =P(Z<-2) = 0.0228 P(X<640) = P(Z<(X-)/)=P(Z<(640-500)/70) =P(Z< 2) = 0.9772 probability that students scored between 360 and 640 is nearlyequal to ( 0.9772-0.0228) = 0.9544 hence no. of students who scored between 360 and 640 =3000 x 0.9544=2863.2 ˜2863
b) A score above 640 is considered exceptional on this exam. Abouthow many students made exceptional scores? P(X>640) = P(Z>(X-)/)=P(Z>(640-500)/70) =P(Z> 2) = 1- 0.9772 = 0.0228 no. of students who made exceptional scores = 0.0228 x 3000 = 68.4 ˜ 68
c) About how many more students received scores between 570 and 640than between 290 and 360? P(X<570) = P(Z<(X-)/)=P(Z<(570-500)/70) =P(Z< 1) = 0.8413 P(X<640) = P(Z<(X-)/)=P(Z<(640-500)/70) =P(Z< 2) = 0.9772 probability that students received scores between 570 and 640= 0.9772-0.8413 = 0.1359 no. of students who received scores between 570 and 640 =0.1359 x 3000 =407.7 ˜407 P(X<290) = P(Z<(X-)/)=P(Z<(290-500)/70) =P(Z<-3) = 0.0013 P(X<360) = P(Z<(X-)/)=P(Z<(360-500)/70) =P(Z<-2) = 0.0228 probability that students received scores between 290 and 360=0.0228-0.0013= 0.0215 no. of students who received scores between 290 and 360=0.0215 x 3000 = 64.5˜ 64 hence no. of more students received scores between 570 and 640than between 290 and 360 = 407-64 = 343
d) If a student received a score of 600, at approximately whatpercentile did she score? P(X=600)=P(Z=(X-)/) =P(Z=(600-500)/70) =P(Z=1.43)=0 .9236 hencea student received a score of 600, approximatelypercentile she scored = 92.36% P(X<640) = P(Z<(X-)/)=P(Z<(640-500)/70) =P(Z< 2) = 0.9772 probability that students scored between 360 and 640 is nearlyequal to ( 0.9772-0.0228) = 0.9544 hence no. of students who scored between 360 and 640 =3000 x 0.9544=2863.2 ˜2863
b) A score above 640 is considered exceptional on this exam. Abouthow many students made exceptional scores? P(X>640) = P(Z>(X-)/)=P(Z>(640-500)/70) =P(Z> 2) = 1- 0.9772 = 0.0228 no. of students who made exceptional scores = 0.0228 x 3000 = 68.4 ˜ 68
c) About how many more students received scores between 570 and 640than between 290 and 360? P(X<570) = P(Z<(X-)/)=P(Z<(570-500)/70) =P(Z< 1) = 0.8413 P(X<640) = P(Z<(X-)/)=P(Z<(640-500)/70) =P(Z< 2) = 0.9772 probability that students received scores between 570 and 640= 0.9772-0.8413 = 0.1359 no. of students who received scores between 570 and 640 =0.1359 x 3000 =407.7 ˜407 P(X<570) = P(Z<(X-)/)=P(Z<(570-500)/70) =P(Z< 1) = 0.8413 P(X<570) = P(Z<(X-)/)=P(Z<(570-500)/70) =P(Z< 1) = 0.8413 P(X<640) = P(Z<(X-)/)=P(Z<(640-500)/70) =P(Z< 2) = 0.9772 probability that students received scores between 570 and 640= 0.9772-0.8413 = 0.1359 no. of students who received scores between 570 and 640 =0.1359 x 3000 =407.7 ˜407 P(X<290) = P(Z<(X-)/)=P(Z<(290-500)/70) =P(Z<-3) = 0.0013 P(X<290) = P(Z<(X-)/)=P(Z<(290-500)/70) =P(Z<-3) = 0.0013 P(X<360) = P(Z<(X-)/)=P(Z<(360-500)/70) =P(Z<-2) = 0.0228 probability that students received scores between 290 and 360=0.0228-0.0013= 0.0215 no. of students who received scores between 290 and 360=0.0215 x 3000 = 64.5˜ 64 no. of students who received scores between 290 and 360=0.0215 x 3000 = 64.5˜ 64 hence no. of more students received scores between 570 and 640than between 290 and 360 = 407-64 = 343
d) If a student received a score of 600, at approximately whatpercentile did she score? P(X=600)=P(Z=(X-)/) =P(Z=(600-500)/70) =P(Z=1.43)=0 .9236 hencea student received a score of 600, approximatelypercentile she scored = 92.36%

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