Baruch College PAF 3401 Homework: Normal Distribution and Probability

This document presents a comprehensive solution to a statistics homework assignment (PAF 3401) focusing on the application of normal distribution. The assignment involves analyzing the first-year salaries of Baruch graduates and the heights of Baruch students. The solution calculates the proportion of students earning below a certain salary, the probability of earning above a certain salary, and the percentage of students earning within a specific range. It also determines the lowest salary for the top 15% and the maximum salary for the first quartile. Furthermore, the solution addresses height-related problems, including finding the height of a student in the top 1%, the height range between the third and fourth quintiles, the proportion of students within a height range, and the height of a student in the first decile. The calculations utilize Z-scores and percentile analysis to provide accurate statistical insights. The document offers a detailed step-by-step approach to solve each problem, making it a valuable resource for students studying statistics and probability.

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PAF 3401
Home Work
Date: 10th March, 2020
1. Normal distribution with mean 50000 and standard deviation 15000
a. Proportion earning less than 40000
The Z= x −m
sd = 40000−50000
15000 =−0.6666
Using the Z score values this gives a cumulative value of 0.2514
Hence we can conclude that 25.14% of the students earn less than $40,000.
b. Probability of a student earning more than 80,000.
Z= 80000−50000
15000 =2
the percentage of students earning less than 80000 from the z score values is
97.72%. . this shows that the percentage that earns over $80,000 is
1−0.9772=2.28 %
c. Earn between 35000 and 65000
earn less than 35000
Z=35000−50000
15000 =−1
probability from the Z score table is ¿ 0.1587
earn less than 65000
Z= 65000−50000
15000 =1
Probability from the table is
¿ 0. 8413
hence the proportion earning between $35000 and $ 65000 is
0.8413−0.1587=0.68.26
which is 68.26%
d. Lowest salary for those in top 15%
First let’s find 100−15=85 %

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Suppose Z= x −50000
15000 =1.0 4
then x= ( 1.04∗15000 ) +50000=65600
Hence the lowest salary for those in top 15% is $ 65,600
e. Max salary for those in 1st quartile
1st quartile is 25%
Now Z= x −50000
15000 =−0.675
x= (−0.675∗15000 ) +50000=39875
Max salary is $39,875
2. Normal distribution with mean 68 and SD 2.5
a. height of student in top 1%
The student at the top 1% is the one occupying the 100th percentile
hence Z= x −68
2.5 =3.5
hence x= ( 3.5∗2.5 ) +68=76.75inches. This is the maximum height
b. Height range between 3rd and 4th quantile
Third quantile is the 75th percentile
z= x−68
2.5 =0.675
x= ( 0.675∗2.5 ) +68=69.69
4th quantile is the 100th percentile which is the maximum value
hence the height range is between 69.69 inches and 76.75 inches
c. proportion between 60 and 65 inches
Less than 60
Z= 60−68
2.5 =−3.2
From the z score table, the value is 0.0007
Proportion less than 65
Z= 65−68
2.5 =−1.2
from the table this gives 0.1151

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Hence the proportion if students between 60 and 65 inches is
0.1151−0.0007=0.1144 equals¿ 11.44 %
d. First decile
This is the 10th percentile
Z= x −68
2.5 =−1.28
x= ( 2.5∗−1 .28 ) +68=64.8 inches
This is the height of the student that falls in the 10th percentile of the data and is nit
the maximum nor the minimum.
e. Students between 65.5 and 70.5 inches
Less than 65.5
Z= 65.5−68
2.5 =−1
this gives 0.1587
less than 70.5
Z=70.5−65.5
2.5 =1
from the table this is 0.8413
Hence the answer is 0.8413−0.1587=0.6826 which is68.26 % of the students.