Summary STK110-TUTORIAL 7B

STK110 Preparation sheet: TUT 7B memo 2023

Please note: When applicable, please round to four (4) decimal places

Question 1 is based on the following information:

It is known that 72% of STK110 students have uncapped internet access. Consider a random sample of
20 STK110 students.

Let 𝑥 = the number of STK110 students with uncapped internet access.
𝑥
𝑝̅ = 𝑛 = the proportion of STK110 students with uncapped internet access.

Use the following extract from Excel to answer questions a to e:

Formula sheet Value sheet




Take note: Rows are hidden for 𝑥 ≤ 6 and 𝑥 ≥ 17.



a. Calculate the total number of experimental outcomes.
220 = 1048576
b. Give the total number of experimental outcomes that with result in 7 STK110 students with
uncapped internet access.
𝑪𝟐𝟎
𝟕 = 𝟕𝟕𝟓𝟐𝟎


c. Calculate the probability that only the last two STK110 students will have no uncapped internet
access (5 dec)
𝑃(𝑌, 𝑌, 𝑌, 𝑌, 𝑌, 𝑌, 𝑌, 𝑌, 𝑌, 𝑌, 𝑌, 𝑌, 𝑌, 𝑌, 𝑌, 𝑌, 𝑌, 𝑌, 𝑁, 𝑁) = 0.7218 0.282 = 0.00021
d. Calculate the probability that exactly 9 of the STK110 students in the random sample have no
uncapped internet access.
20
𝑃(𝑥 = 11) = ( ) (0.28)9 (0.82)11 = 0.0479
11
e. Determine the probability that more than 10 of the STK110 students in the random sample have no
uncapped internet access.
Means 11 or more do not have uncapped.
If 11 or more do not have uncapped, then 9 or less do have uncapped.
𝑃(𝑥 ≤ 9) = 0.0100
f. Calculate the probability that the proportion of STK110 students with uncapped internet access is
more than 0.85.
0.85 − 0.72
𝑃(𝑝̅ > 0.85) = 𝑃 (𝑧 > ) = 𝑃(𝑧 > 1.29) = 1 − 0.9015 = 0.0985
0.1004