Question

Instructions This assignment is to be typed up in the supplied R-Script. You need to show all of your work in R in the given script.

3. Infant mortality. The infant mortality rate is defined as the number of infant deaths per 1,000 live births. This rate is often used as an indicator of the level of health in a country. The relative frequency histogram below shows the distribution of estimated infant death rates for 224 countries for which such data were available in 2014.

(a) Estimate Q1, the median, and Q3 from the histogram.

(b) Would you expect the mean of this data set to be smaller or larger than the median? Explain your reasoning.

(c) If you calculated the z-score for the median in this distribution, would the result be positive or negative? Explain your reasoning.

Answer 1

mid <- seq(5,125,10)
f <- c(0.35,0.25,0.125,0.02,0.075,0.06,0.05,0.04,0.005,0.01,0.01,0.002,0.003)
pmfTable <- cbind.data.frame(mid,f)
pmfTable$cumFreq = cumsum(pmfTable$f)

Q1 = N+1/ 4 th rank = 1000+1 / 4th rank =250.25th rank. This is less than C.F. 350. 1st class is Q1 class. Q1 value is lies between 0 and 10

Median =Q2 = 2(N+1)/ 4 th rank = 2(1000+1) / 4th rank =500.5th rank. This is less than C.F. 600. 2nd class is median class. Median value is lies between 10 and 15

Q3 = 3(N+1)/ 4 th rank = 3(1000+1) / 4th rank =750.75th rank. This is less than C.F. 820. 5th class is Q3 class. Q3 value is lies between 40 and 50.

(b) Mean = sum(mid value * frequencies) / Total frequency

              = (350*5 + 250*15 + 125*25+ ....+125*3)/1000 = 24.855

Here Mean value is greater than Median values since the shape of the histogram is postive skewed distribution