Question

Tree heights: Cherry trees in a certain orchard have heights that are normally distributed with mean = 117 inches and standard deviation o = 14 inches. Use the TI-84 PLUS calculator to answer the following. Round the answers to at least four decimal places. (a) What proportion of trees are more than 129 inches tall? (b) What proportion of trees are less than 100 inches tall? (c) What is the probability that a randomly chosen tree is between 85 and 100 inches tall? Part: 0/3 Part 1 of 3 What proportion of trees are more than 129 inches tall? The proportion of trees that are more than 129 inches tall is :

Answer 1

Using TI84

Part a)

X ~ N ( µ = 117 , σ = 14 )
P ( X > 129 ) = 1 - P ( X < 129 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 129 - 117 ) / 14
Z = 0.8571
P ( ( X - µ ) / σ ) > ( 129 - 117 ) / 14 )
P ( Z > 0.8571 )
P ( X > 129 ) = 1 - P ( Z < 0.8571 )
P ( X > 129 ) = 1 - 0.8043
P ( X > 129 ) = 0.1957

Press button 2ND + VARS

ISTR DRAW 1 C normalPdf 28normalcdf( 3: invNorm 4: invT 5:tpdf 6:tcdf 7JPdf NM455r

Select normalcdf

Putting the information

normaicdf lower: 129 UPPER: 1E99 : 117 a:14 Paste

Will get the probability 0.1957

normalcdf(129,1 .1956829201

Part b)

X ~ N ( µ = 117 , σ = 14 )
P ( X < 100 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 100 - 117 ) / 14
Z = -1.2143
P ( ( X - µ ) / σ ) < ( 100 - 117 ) / 14 )
P ( X < 100 ) = P ( Z < -1.2143 )
P ( X < 100 ) = 0.1123

Press button 2ND + VARS

ISTR DRAW 1 C normalPdf 28normalcdf( 3: invNorm 4: invT 5:tpdf 6:tcdf 7JPdf NM455r

Select   normalcdf

Putting the information

OFMA1EdF nOrMaicdf lower: -1E99 UPPer: 100 :117 T: 14 Paste

normalcdf( -1 E99 .1123193824

Will get the probability 0.1123

Part c)

X ~ N ( µ = 117 , σ = 14 )
P ( 85 < X < 100 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 85 - 117 ) / 14
Z = -2.2857
Z = ( 100 - 117 ) / 14
Z = -1.2143
P ( -2.29 < Z < -1.21 )
P ( 85 < X < 100 ) = P ( Z < -1.21 ) - P ( Z < -2.29 )
P ( 85 < X < 100 ) = 0.1123 - 0.0111
P ( 85 < X < 100 ) = 0.1012

Press button 2ND + VARS

ISTR DRAW 1 C normalPdf 28normalcdf( 3: invNorm 4: invT 5:tpdf 6:tcdf 7JPdf NM455r

Select   normalcdf

Putting the information

normaicdf lower: 85 UPPer: 100 :117 14 T: Paste

Will get the probability  0.1012

normalcdf(85, 10 .1011839249