Sweden is reported to have among the highest percentage of citizens aged 80 years and over, at approximately 5.5 % (i.e. 5.5 % of it's population falls within this age category). Suppose 20 individuals are randomly selected from the population, and X is the number of individuals out of 20 that are 80 or more years old.
a) On average, how many of the 20 individuals would be 80 years old or older?
E(x)=np= 20*0.055= 1.1
is this corrected?
b)What is the probability that less than 2 of the 20 individuals are 80 years old or older?
P(x<2)=P(x=0)+P(x=1)= 0.6980679
is this corrected?
Looks good to me! Good job!
a) Yes, you are correct. To calculate the average number of individuals who are 80 years old or older, you can use the formula E(X) = np, where n is the total number of individuals (20 in this case) and p is the probability of an individual being 80 or older. Since the probability is 0.055, the calculation would be E(X) = 20 * 0.055 = 1.1.
b) To calculate the probability that less than 2 out of the 20 individuals are 80 years old or older, you need to add the probabilities of having exactly 0 or 1 individual in that age category. Using the binomial probability formula, P(X < 2) = P(X = 0) + P(X = 1).
P(X = 0) = (20 choose 0) * (0.055^0) * (0.945^20) = 0.6980679
P(X = 1) = (20 choose 1) * (0.055^1) * (0.945^19) = 0.2781312
Therefore, P(X < 2) = 0.6980679 + 0.2781312 = 0.9761991.
So the probability that less than 2 out of the 20 individuals are 80 years old or older is approximately 0.9761991.