Question

unit 3 topic 12 excercise 12-14

-yrLua Rossman, Workshop Statistics Discovery with Data, te States (11511 , Study Practice Assignment Gradebook Open Assignment TIHEN RESOURCES Dog Heights Suppose that heights of male German Shepherds follow a normal distribution with mean 23 inches and standard deviation 2.5 inches. Suppose further that heights of male Shelties follow a normal distribution with mean 16 inches and standard deviation 1.5 inches a. What percentage of male German Shepherds have a height between 20.5 and 25.5 inches? Round your answer to two decimal places. % b. Between what two values do roughly 95% of male Sheltie's heights fall? Enter your answers in increasing order, About 95% of all male Sheltie's heights fall between inches and inches. C. How tall must a male Sheltie be to be among the tallest 10% of Shelties? Round your answer to two decimal places. inches d. Which is more unusual: a male German Shepherd having a heiaht of greater than 26 inches or a male Sheltie havina a heiaht of greater than 19 PoliorI 2000-2020 John Wiley & Sons, Inc. All Rights Reserved. A Division of John Wiley & Sons, Inc.

Answer 1

a)

µ = 23, σ = 2.5

P(20.5 < X < 25.5) =

= P( (20.5-23)/2.5 < (X-µ)/σ < (25.5-23)/2.5 )

= P(-1 < z < 1)

= P(z < 1) - P(z < -1)

Using excel function:

= NORM.S.DIST(1, 1) - NORM.S.DIST(-1, 1)

= 0.6827 = 68.27%

b)

µ = 23, σ = 2.5

Proportion in the middle = 0.95

Proportion on left and right side of normal curve = 0.025

Z score at p = 0.025 using excel = NORM.S.INV(0.025) = 1.96

Value of X1 = µ - z*σ = 23 - (1.96)*2.5 = 18.10

Value of X2 = µ + z*σ = 23 + (1.96)*2.5 = 27.90

c)

µ = 16, σ = 1.5

P(x > a) = 0.1

= 1 - P(x < a) = 0.1

= P(x < a) = 0.9

Z score at p = 0.9 using excel = NORM.S.INV(0.9) = 1.2816

Value of X = µ + z*σ = 16 + (1.2816)*1.5 = 17.92

d)

For Germen Shepherd:

µ = 23, σ = 2.5

P(X > 26) =

= P( (X-µ)/σ > (26-23)/2.5)

= P(z > 1.2)

= 1 - P(z < 1.2)

Using excel function:

= 1 - NORM.S.DIST(1.2, 1)

= 0.1151

For sheltie:

µ = 16, σ = 1.5

P(19 < X < 25.5) =

= P( (19-16)/1.5 < (X-µ)/σ < (25.5-16)/1.5 )

= P(2 < z < 6.3333)

= P(z < 6.3333) - P(z < 2)

Using excel function:

= NORM.S.DIST(6.3333, 1) - NORM.S.DIST(2, 1)

= 0.0228

Therefore, Sheltie is more unusual.