TU Wien:Statistik und Wahrscheinlichkeitstheorie UE (Levajkovic)/Übungen 2023W/HW05.3

Human resource testing

Some human resource deprtments administer standard IQ tests to all employees. The Stanford- Binet test scores are well modeled by a Normal model with expectation 100 and standard deviation 16.

(a)  If the applicant pool is well modeled by this distribution, what is the probability that a randomly selected applicant would have the score ibetween 84 and 116?

(b)  For the IQ test administered by human resources, what cutoff value would separate the middle 95%?

Dieses Beispiel ist als solved markiert. Ist dies falsch oder ungenau? Aktualisiere den Lösungsstatus (Details: Vorlage:Beispiel)

Lösungsvorschlag von Friday

--Friday Sa 30 Jan 2021 16:56:21 CET

# Statistics and Probability HW #5
# Friday 
# Duedate: 09.11.2020

# Problem 1 - Human resource testing
# Some human resource departments administer standard IQ tests to all employees.
# The Stanford-Binet test scores are well modeled by a Normal model with 
# expectation 100 and standard deviation 16.
ex = 100
sd = 16

# Problem 1a)
# If the applicant pool is well modeled by this distribution, a randomly 
# selected applicant would have what probability of scoring in the region 
# between 84 and 116?
result_1a <- pnorm(116, mean=ex, sd=sd) - pnorm(84, mean=ex, sd=sd)
result_1a

# Problem 1b)
# For the IQ test administered by human resources, what cutoff value would 
# separate the middle 95%?
result_1b <- c(qnorm(0.025, mean=ex, sd=sd), qnorm(0.95 + 0.025, mean=ex, sd=sd))
result_1b

Lösungsvorschlag von Simplex

# a)
expectation <- 100
sd <- 16

res1a <- pnorm(116, expectation, sd) - pnorm(84, expectation, sd)
print(res1a)

# b)
res1b_lower <- qnorm(0.025, expectation, sd)
res1b_upper <- qnorm(0.975, expectation, sd)

print(c(res1b_lower, res1b_upper))

(a)

$${\mathcal {N}}(100,16^{2}):P(84<X\leq 116)=P(X\leq 116)-P(84\leq X)\approx 68\%{\text{ (represents }}\pm {\text{ one sd)}}$$

(b)

$$\Phi ^{-1}(0.025)\cdot \underbrace {16} _{\sigma }+\underbrace {100} _{\mu }=68.64{\text{ (Lookup cell in table and read axis)}}$$

$$\Phi ^{-1}(0.975)\cdot \underbrace {16} _{\sigma }+\underbrace {100} _{\mu }=131.36{\text{ (Lookup cell in table and read axis)}}$$

--Simplex 13:43, 3. Feb. 2023 (CET)