Question

1. the weight of an average nfl lineman in 2024 is 315 lbs. with a standard deviation of 20. x = the weight of a randomly selected lineman. find:

a. p(x < 300)
b. p(x > 350)
c. p(280 < x < 305)
d. p(x = 300)
e. p(270 < x < 290)

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 315\) (mean) and \(\sigma=20\) (standard deviation).

Part A

For \(x = 300\), \(z=\frac{300 - 315}{20}=\frac{-15}{20}=-0.75\)
Using the standard normal table, \(P(X < 300)=P(Z<-0.75)\)
From the standard normal table, \(P(Z < - 0.75)=0.2266\)

Part B

For \(x = 350\), \(z=\frac{350 - 315}{20}=\frac{35}{20}=1.75\)
\(P(X>350)=1 - P(X\leqslant350)\)
\(P(X\leqslant350)=P(Z\leqslant1.75)\)
From the standard normal table, \(P(Z\leqslant1.75) = 0.9599\)
\(P(X>350)=1 - 0.9599=0.0401\)

Part C

For \(x = 280\), \(z=\frac{280 - 315}{20}=\frac{-35}{20}=-1.75\)
For \(x = 305\), \(z=\frac{305 - 315}{20}=\frac{-10}{20}=-0.5\)
\(P(280<X<305)=P(-1.75<Z<-0.5)\)
\(P(-1.75<Z<-0.5)=P(Z < - 0.5)-P(Z < - 1.75)\)
From the standard normal table, \(P(Z < - 0.5)=0.3085\) and \(P(Z < - 1.75)=0.0401\)
\(P(-1.75<Z<-0.5)=0.3085 - 0.0401=0.2684\)

Part D

Since \(X\) is a continuous random variable (weight is a continuous quantity), \(P(X = 300)=0\)

Part E

For \(x = 270\), \(z=\frac{270 - 315}{20}=\frac{-45}{20}=-2.25\)
For \(x = 290\), \(z=\frac{290 - 315}{20}=\frac{-25}{20}=-1.25\)
\(P(270<X<290)=P(-2.25<Z<-1.25)\)
\(P(-2.25<Z<-1.25)=P(Z < - 1.25)-P(Z < - 2.25)\)
From the standard normal table, \(P(Z < - 1.25)=0.1056\) and \(P(Z < - 2.25)=0.0122\)
\(P(-2.25<Z<-1.25)=0.1056-0.0122 = 0.0934\)

Answer:

A. \(0.2266\)
B. \(0.0401\)
C. \(0.2684\)
D. \(0\)
E. \(0.0934\)