Code and results

The above "prediction using the normal distribution" was performed, and the accuracy was compared with the "prediction that the minimum passing score = (average score of examinees + average score of successful applicants) / 2".

import numpy as np
from scipy.stats import norm
import matplotlib.pyplot as plt
from sklearn.metrics import r2_score, mean_absolute_error

# 2013-2019 data from https://www.goro-goro.net/2019-musashi
jukensya = [433, 556, 519, 590, 577, 541, 569]
goukakusya = [177, 177, 185, 183, 187, 185, 186]
juken_heikin = [138.5, 173.3, 172.9, 167.0, 165.5, 186.9, 170.5]
goukaku_heikin = [166.5, 210.7, 210.3, 202.0, 197.0, 221.5, 205.2]
goukaku_saitei = [146, 192, 188, 184, 180, 201, 185]

goukaku_saitei_pred = []

for i in range(7):  # 2013-Analysis of 7 times in 2019
    r = jukensya[i] / goukakusya[i]  #magnification
    mu_a = juken_heikin[i]  #Candidate average score
    mu_p = goukaku_heikin[i]  #Average score of successful applicants
    
    sigma_l = 0.1
    sigma_r = 1000
    sigma = (sigma_l + sigma_r) / 2  #Standard deviation, 0.Search in the range of 1 to 1000
    
    for i in range(100):  #Repeat the dichotomy 100 times
        b = norm.isf(1 / r, mu_a, sigma)  #Provisional minimum passing score
        mu_p_prime = mu_a + sigma * norm.pdf((b - mu_a) / sigma) \
                     / (1 - norm.cdf((b - mu_a) / sigma))  #Provisional average score of successful applicants
        if mu_p_prime < mu_p:
            sigma_l = sigma
        else:
            sigma_r = sigma
        sigma = (sigma_l + sigma_r) / 2
    
    goukaku_saitei_pred.append(b)

#Prediction that the midpoint between the average score of examinees and the average score of successful applicants is the lowest passing score
goukaku_saitei_pred_rough = [(goukaku_heikin[i] + juken_heikin[i]) / 2 for i in range(7)]

## R^Confirmation of 2 and MAE##

print("Pred 1:Results of prediction using normal distribution")
print("R^2: {:.4f}".format(r2_score(goukaku_saitei, goukaku_saitei_pred)))
print("MAE: {:.4f}".format(mean_absolute_error(goukaku_saitei, goukaku_saitei_pred)))
print("")
print("Pred 2: (Candidate average score+Average score of successful applicants) /The result of the prediction that it is 2")
print("R^2: {:.4f}".format(r2_score(goukaku_saitei, goukaku_saitei_pred_rough)))
print("MAE: {:.4f}".format(mean_absolute_error(goukaku_saitei, goukaku_saitei_pred_rough)))

##Measured value-Creating a forecast plot##

fig = plt.figure(figsize=(10, 5))
lim = [140, 220]  #Range of both axes
s = 17  #font size

# Pred 1:Results of prediction using normal distribution
ax0 = fig.add_subplot(1,2,1)

ax0.plot(goukaku_saitei, goukaku_saitei_pred, "o", markersize=8)
ax0.plot(lim, lim, "k-")

ax0.set_title('Pred 1', fontsize=s)
ax0.set_xlabel('True', fontsize=s)
ax0.set_ylabel('Predicted', fontsize=s)
ax0.tick_params(labelsize=s)
ax0.set_xlim(lim)
ax0.set_ylim(lim)
ax0.set_aspect('equal')

# Pred 2: (Candidate average score+Average score of successful applicants) /The result of the prediction that it is 2
ax1 = fig.add_subplot(1,2,2)

ax1.plot(goukaku_saitei, goukaku_saitei_pred_rough, "o", markersize=8)
ax1.plot(lim, lim, "k-")

ax1.set_title('Pred 2', fontsize=s)
ax1.set_xlabel('True', fontsize=s)
ax1.set_ylabel('Predicted', fontsize=s)
ax1.tick_params(labelsize=s)
ax1.set_xlim(lim)
ax1.set_ylim(lim)
ax1.set_aspect('equal')

#Save as PNG
fig.tight_layout()
fig.savefig("plot.png ")

output The R ^ 2 value is the coefficient of determination, and the closer it is to 1, the better the accuracy. MAE is the average of the absolute values of the prediction errors, and the closer it is to 0, the better the accuracy.

Pred 1:Results of prediction using normal distribution
R^2: 0.9892
MAE: 1.4670

Pred 2: (Candidate average score+Average score of successful applicants) /The result of the prediction that it is 2
R^2: 0.9583
MAE: 2.5571

Consideration

It can be said that Pred 1, which is a prediction using a normal distribution, is more accurate </ b> than Pred 2, which is a simple prediction. The average prediction error is about 1.5 points, which seems to be sufficient as a guide for actual study. Also, looking at the plot of Pred 2, the prediction error is large in the data near 145 points in the lower left. This is probably because Pred 2 uses the "midpoint between the average score of examinees and the average score of successful applicants" regardless of the exam magnification, so it cannot respond to changes in the magnification (only this year). The magnification is as low as 2.4 times, 2.8 to 3.2 times in other years). On the other hand, in Pred 1, the prediction error is suppressed to the same level as other data or less, and it can be said that the method using the normal distribution can cope with the change in magnification.