Question

1. From a random sports news, the probability of observing the word “ball” and “player” is 0.8 and 0.7, respectively. For a non-sports news, the probability to observe “ball” is 0.1, so does that to observe “player”. Let’s assume that in any article, the appearance of any two words (including “ball” and “player”) are independent with each other. Also, the probability of sports news’ occurrences is 0.2. Given a news report x containing both “ball” and “player”, what is the probability that x is a sports news.

2. To estimate the coronavirus infection situation in a city (with large population), the researchers randomly sample N people therein and test their body temperatures.

(a) If N = 5 and the tested results are 36.2, 38.9, 39.2, 38.4, 36.8 ( ◦ C). Calculate the sample mean, variance, standard deviation, medium, and range.
(b) If we want to propose a statistical model (i.e., a probability distribution) to reflect people’s average temperature in this city. How will you design the model? How to approximate the parameters in the model?
(c) As per a previous study, human body temperatures, denoted as X, satisfy normal distribution, where the mean is 37◦ C and standard deviation 0.4◦ C, i.e., X ∼ N (37, 0.4). Suppose that we test the body temperatures for 16 people in the city whose sample mean turned out to be 37.3◦ C. Let’s use µ to denote the expected value of people’s body temperatures in this city and further make a null hypothesis H0: µ = 37◦ C. Will you accept H0 at significance level 5% and 10%, respectively?

Answer 1

1:

Let S shows the event of sports news and N shows the event of non-sports news. So we have

P(S) = 0.20, P(N) = 1 - P(S) = 0.80

Let B shows the event of observing word ball and P shows the event of observing word player. So,

P(B |S) = 0.80, P(P | S) = 0.70

P(B | N) = 0.10, P(P | N) = 0.10

Since oberving each word is independent from other so

P(B and P |S) = P(B|S)P(P|S) = 0.80 *0.70 = 0.56

P(B and P |N) = P(B|N)P(P|N) = 0.10 *0.10 = 0.01

The probability that x is a sports news given a news report x containing both “ball” and “player” is

P(S | B and P) = [ P(B and P | S) P(S) ] / [ P(B and P | S) P(S) + P(B and P | N) P(N) ] = [ 0.56 * 0.2 ] / [ 0.56 *0.2 + 0.01 * 0.80 ] = 0.9333

Answer: 0.9333