Probability Theory and Statistics Homework: Error, IQ & Uniform Dist

This document presents solutions to a statistics homework assignment focusing on probability theory. It addresses several questions, including calculating probabilities of Type 1 errors in a series of statistical tests using binomial distribution, determining probabilities related to missing stickers in a Panini album, analyzing IQ scores using normal distribution, examining probabilities in an exam with a uniform distribution of results, and calculating employee statistics related to the number of children. The solutions involve applying statistical formulas and concepts such as z-scores, binomial distribution, and understanding measurement scales for quantitative data. The document concludes by identifying the ratio scale as the most suitable measurement scale for the employee data.

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Question 1
Probability of Type 1 Error = 0.05
Probability of no type 1 error = 1-0.05 = 0.95
Number of trials = 5
The given distribution is an example of binomial distribution since there are only two
outcomes and outcome in each trial is independent of the outcome in previous trial.
a) P(0) = 5C0*(0.05)0*(0.95)5 = 0.7738
b) P(1) = 5C1*(0.05)1*(0.95)4 = 0.2036
c) P(5) = 5C5*(0.05)5*(0.95)0 = 0.0000
d) At most once = P(0) + P(1) = 0.7738 + 2036 = 0.9774
Question 2
a) Requisite probability = P(0) = (645/650)5 = 0.9621
b) Required probability = P(1) = 5C1*(5/650)*(645/650)4 = 0.0373
c) Required probability = 1- P(0) = 1-0.9621 = 0.0379
d) Required probability = P(5) = (5/650)*(4/650)*(3/650)*(2/650)*(1/650) = 0.0000
Question 3
a) The formula for z score is indicated below.
Z = (X-μ)/σ
Here, X = 130, μ = 100 and σ = 15
Hence, Z = (130-100)/15 = 2
P (Z≤2) = 0.9772 (Obtained from z table)

Therefore, P(Z>2) = 1- P (Z≤2) = 1-0.9772 = 0.02275
b) The formula for z score is indicated below.
Z = (X-μ)/σ
Here, X = 75, μ = 100 and σ = 15
Hence, Z = (75-100)/15 = -1.67
P(Z<-1.67) = 0.0478 (Obtained from z table)
c) Since 95% of the population needs to be captured around μ, hence on each side 47.5% of
the values ought to be captured. The z score associated with the point on the left side of the
mean and the right side of the mean are -1.96 and 1.96.
Hence, lower end of the IQ score = 100 – 1.96*15 = 70.6
Upper end of the IQ score = 100 + 1.96*15 = 129.4
Thus, the desired IQ interval lies between 70.6 and 129.4 with a width of 58.8.
Question 4
It is known that the exam results tend to follow a uniform distribution which implies that the
score distribution is uniform across the marks spectrum.
a) Students scoring 50% or above would be 50% owing to uniform distribution of marks
across the students. Hence, probability that a student passes the exam = ½ = 0.5
b) Owing to a uniform distribution, 1% of the students would be stored above 99%. Further
between 99% and 100%, there are 100 sub intervals where the marks distribution would be
uniform.
Hence, requisite probability for a perfect score = (1/100) = 0.01
c) 80% of the students would have not exceeded 80% score and only 20% would have
exceeded the same.
Hence, requisite probability = 0.2

Question 5
a) Total number of employees = 50
Employees that have no kids or 0 kids = 9
Hence, share of such employees = (9/50) = 0.18 or 18%
b) Total number of employees = 50
Employees that have 2 or more kids = 29
Hence, share of such employees = (29/50) = 0.58 or 58%
c) The given data is quantitative or numerical in nature and the suitable measurement scale
for the data used here is ratio considering the fact that absolute zero can be defined as
negative value for children is not possible.