Question

Jimmy is the education consultant of a tutorial school, which arranges preparatory classes for IELT examination. He randomly checks the mock examination result of 22 students in the school and the results are as follow:
7.9 8.2 7.8 5.4 6.8 6.8 7.1 7.2 7.9 8.1 7.6 7.8 5.8 6.3 7.7 7.7 8.3 8.3 4.7 5.1 7.3 6.5
(a) Write a report to reflect students’ ability in the mock examination.
average performance, the median, the standard deviation, the bottom 20% and top 10% performance level.
(b) The manager asks if the distribution of the examination result looks like a normal distribution. Answer his question by reviewing the skewness of the data with detailed calculation.
(c) There are 18 students in Alex’s class. By using the sampled data as a reference to project the chance an individual student can get more than 6 marks and assume students’ result are independent, calculate the probability that more than 80% of Alex’s students can score higher than 6 marks in the mock examination.

Q: a-c answer

Answer 1

a) Using Excel we find mean, median, standard deviation. Using PERCENTILE.INC () we find the bottom 20% and top 10% performance level.

b) We build a histogram. From the histogram we can see that the distributiion is left-skewed (left tail is longer). Skewness equals -0.95. So the distribution is negatively skewed (mean<median). For normal distribution skewness is equal to 0. Kurtosis (whether the data are heavy-tailed or light-tailed) equals -0.07. We see that kurtosis does not differ much from 0. For standard normal distribution skewness is equal to 0. The distribution of the examination result does not look like a normal distribution.

c) $n=18, N=22, P=\frac{18}{22}\text{ is population proportion}.$

$\mu_{\overline{p}}=P=\frac{18}{22}.\\ \frac{n}{N}>0.05.\text{ So we have: }\\ \sigma_{\overline{p}}=\sqrt{\frac{P(1-P)}{n}}\sqrt{\frac{N-n}{N-1}}=\frac{2}{11}\sqrt{\frac{1}{21}}.\\ P\{\overline{p}>0.8\}=1-P\{\overline{p}\leq 0.8\}=1-F(0.8)=\\ =1-\Phi(\frac{0.8-\frac{18}{22}}{\frac{2}{11}\sqrt{\frac{1}{21}}})\approx 1-0.3234=0.6766.\\$