Question

TIMSS 2003 (Trends in International Mathematics and Science Study) focused on the mathematics and science achievement of eighth-graders throughout the world. A total of 45 countries (including the United States) participated in the study. The mean math exam score for U.S. students was 513 with a standard deviation of 93. Assume the scores are normally distributed and a sample of 157 students is taken. (Give your answers correct to four decimal places.)

(a) Find the probability that the mean TIMSS score for a randomly selected group of eigth-graders would be between 496 and 516.


(b) Find the probability that the mean TIMSS score for a randomly selected group of eighth-graders would be less than 524.

Answer 1

X :TIMSS 2003 Math score of eighth-graders

X follows normal distribution with mean $\mu$ = 513 and standard deviation $\sigma$ = 93

Sample size : n= 157

Sampling distribution of sample mean : $\overline{x}$ follows normal with mean $\mu _{\overline{x}}$ = 513 and standard deviation:

ở Vn = 93,157 = 7.42

(a) Probability that the mean TIMSS score for a randomly selected group of eigth-graders would be between 496 and 516 =  

P(496 < 5 < 516)

P(496 <7 < 516) = PT < 516) – PT< 496)

Z-score for 516 = (516-513)/7.42 = 0.40; Z-score for 496 = 496-513/7.42 = -2.29

From standard normal tables,

P(Z $\small \leq$ 0.40) = 0.6554 ; P(Z $\small \leq$ -2.29)=0.0110

PCT 5 516) = P(Z < 0.40) = 0.6554 PT < 496) = P(Z < -2.29) = 0.0110

P(496 < I< 516) = PT < 516) – PT < 496) = 0.6554 -0.0110 = 0.6444

Probability that the mean TIMSS score for a randomly selected group of eigth-graders would be between 496 and 516 =0.6444

(b) Probability that the mean TIMSS score for a randomly selected group of eighth-graders would be less than 524

= PT < 524)

Z-score for 524 = (524-513)/7.42 = 1.48

from standard normal tables

P(Z<1.48) = 0.9306

PT <524) = P(Z < 1.48) = 0.9306

Probability that the mean TIMSS score for a randomly selected group of eighth-graders would be less than 524 =0.9306