Application 1 The scores of a reference population on the Wechsler Intelligence

Question

Application 1

The scores of a reference population on the Wechsler Intelligence Scale for children (WISC) are normally distributed with mean 100 and standard deviation equal 15.

Use theR codes to answer the following:

What percent of this population have WISC scores below 80?

Above 120?

Between 80 and 120?

What score will place a child in the top 5% of the population?

Application 2:

The cholesterol levels for women aged 20 to 34 follow an approximately normal distribution with mean 185 (mg/dl) and standard deviation 39 mg/dl.

What percent of young women have blood cholesterol:

a.     Above 240 mg/dl?

b.     Between 200 and 240 mg/dl?

c.      Less than 220?

d.     Find the cutoff point that corresponds to the top 5% of Cholestrol levels?

Explanation / Answer

Answering Application 1 in detail and have included code for Application 2

I will try to include all information in the R code format:

#Clear environment
rm(list=ls(all=T))
#Mean mu = 100
mu<-100
#Standard deviation, sd = 15
sd<-15

#Qn1:What percent of this population have WISC scores below 80?
#The function pnorm() with score as 80(q=80), mean , sd and lower.tail = TRUE will give us the answer
pnorm(q = 80,mean = mu,sd = sd,lower.tail = TRUE )
#output : 0.09121122
#Hence answer: 9.12% of this population have WISC scores below 80

#Qn2:What percent of this population have WISC scores above 120?
#The function pnorm() with score as 120(q=80), mean , sd and lower.tail = FALSE will give us the answer
pnorm(q = 120,mean = mu,sd = sd,lower.tail = FALSE )
#output : 0.09121122
#Hence answer: 9.12% of of this population have WISC scores above 120

#Qn3:What percent of this population have WISC scores between 80 and 120?
#pnorm() with q = 80 and lower.tail = TRUE gives us the percentage of the population with scores below 80
#and pnorm() with q = 120 and lower.tail = TRUE gives us the percentage of population with scores below 120
#hence the difference of the above two percentages will give us the percenatge of population between 80 and 120
pnorm(q = 120,mean = mu,sd = sd,lower.tail = TRUE) - pnorm(q = 80,mean = mu,sd = sd,lower.tail = TRUE)
#output: 0.8175776
#Answer: 81.75776% of this population have WISC scores between 80 and 120

#Qn4: What score will place a child in the top 5% of the population?
#The score in the top 5% will be given by the following function
#qnorm() with p = 0.05 (5%), mean = mu , sd = sd and lower.tail = FALSE
qnorm(p = 0.05,mean = mu,sd = sd,lower.tail = FALSE)
#Output : 124.6728
#Answer: A score of at least 124.6728 will place the child in the top 5% of the population

#2nd Part applying the same logic as above in answering
#Qn1
pnorm(q=240,mean = 185,sd=39,lower.tail = FALSE)
#[1] 0.07923199 = 7.92%

#Qn2
pnorm(q = 240,mean = 185,sd = 39,lower.tail = TRUE) - pnorm(q = 200,mean = 185,sd = 39,lower.tail = TRUE)
#Output: [1] 0.2710292 = 27.10%

#Qn 3
pnorm(q=220,mean = 185,sd=39,lower.tail = TRUE)
#output: 0.8152568 = 81.53%

#Qn4
qnorm(p=0.05,mean = 185,sd=39,lower.tail = FALSE)
#Output: 249.1493

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