Omid - Challenge 84

data-challenges
advanced-exercises
🔰 In the question table, the areas between 0 and Z of the normal distribution are provided.
Published

March 24, 2026

Illustration for Omid - Challenge 84

Challenge Description

🔰 In the question table, the areas between 0 and Z of the normal distribution are provided.

Solutions

library(tidyverse)
  library(readxl)
  
  path = "files/CH-84 Normal Distribution.xlsx"
  input1 = read_excel(path, range = "B2:L13")
  input2 = read_excel(path, range = "N2:N9")
  test   = read_excel(path, range = "O2:O9")
  
  result1 = input1 %>%
    pivot_longer(cols = -c(1), names_to = "Z1", values_to = "prob") %>%
    mutate(Z1 = as.numeric(Z1), 
           Z_tot = Z1 + Z) %>%
    select(Z = Z_tot, prob)
  
  result2 = tibble(Probability = input2$Probability) %>%
    rowwise() %>%
    mutate(Z = result1 %>%
             mutate(diff = abs(prob - Probability)) %>%
             filter(diff == min(diff)) %>%
             pull(Z)) %>%
    ungroup()
    
  all.equal(result2$Z, test$Z)
  #> [1] TRUE

  • Logic:

    • Reads the workbook ranges needed for the challenge

    • Reshapes the data into the grain required by the task

    • Builds the intermediate columns that drive the final result

  • Strengths:

    • The R solution stays close to the workbook rule and keeps the transformation compact.

  • Areas for Improvement:

    • The code assumes the sheet structure and source ranges remain stable.

  • Gem:

    • The strongest part of the solution is choosing the right intermediate representation before shaping the final output.
import pandas as pd

path = "CH-84 Normal Distribution.xlsx"

input1 = pd.read_excel(path, usecols="B:L", skiprows=1)
input2 = pd.read_excel(path, usecols="N", skiprows=1, nrows=7)
test = pd.read_excel(path, usecols="O", skiprows=1, nrows=7)
test.columns = test.columns.str.replace('.1', '')

result1 = input1.melt(id_vars=["Z"], var_name="Z1", value_name="prob")
result1["Z_tot"] = result1["Z1"] + result1["Z"]
result1 = result1[["Z_tot", "prob"]]

result2 = pd.DataFrame({"Probability": input2["Probability"]})
result2["Z"] = result2["Probability"].apply(lambda x: result1.loc[(result1["prob"] - x).abs().idxmin(), "Z_tot"])

result2["Z"] = result2["Z"].round(2)
test["Z"] = test["Z"].round(2)

print(result2["Z"].equals(test["Z"]))  # True

  • Logic:

    • Reads the workbook ranges needed for the challenge

    • Reshapes the data into the grain required by the task

  • Strengths:

    • The Python version follows the same rule in a direct dataframe-oriented implementation.

  • Areas for Improvement:

    • The code assumes the workbook layout remains stable, so any sheet redesign would require small adjustments.

  • Gem:

    • The implementation stays close to the original workbook rule instead of adding unnecessary abstraction.

Difficulty Level

This task is moderate:

  • The core logic is clear, but the correct transformation pattern is not obvious from the raw input.

  • The challenge combines multiple reshaping, grouping, or parsing steps.