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1. Suppose that the weights of 3500 registered female Great Danes in the United States aredistributed normally with a mean of 125 lb. and a standard deviation of 5.2 lb.
Approximately how many of the Great Danes weigh more than 130.2 lbs.?

2. Body mass index, or BMI, of 18-year-old females are normally distributed with a mean of 24.6 and
a standard deviation of 2.4. Eighteen-year-old Sasha BMI has a z-score of 1.22.
What is Sasha’s BMI?

3. An experiment compares weight of fish for two different brands fish food. Nacho Average Minos
fish food has a mean fish weight 12 lbs., and a standard deviation of 1.4 lbs. Impossible Minos fish
food has a mean fish weight of 14 lbs. and a standard deviation of 1.2 lb. A fish is measured to be
13.1 lbs. Which brand of fish food, Nacho Average Minos, or Impossible Minos, is more likely to have
been used to feed this fish?

4. Height of 10th grade boys is normally distributed with a mean of 66.9 in. and a standard deviation
of 2.5 in.
The area greater than the z-score is the probability that a randomly selected 15-year-old boy
exceeds 71 in.
What is the probability that a randomly selected 10th grade boy exceeds 71 in.?

5. Heights for 16-year-old girls are normally distributed with a mean of 64 in. and a standard
deviation of 1.9 in.
Find the z-score associated with the 90th percentile.
Find the height of a 16-year-old girl in the 90th percentile.

Apr 27, 2022

#1
+4

1. Suppose that the weights of 3500 registered female Great Danes in the United States aredistributed normally with a mean of 125 lb. and a standard deviation of 5.2 lb.
Approximately how many of the Great Danes weigh more than 130.2 lbs.?

(130 - 125)  /   5.2  =   z score  of .96  =   table value of .8315

So

3500  ( 1 -.8315)  ≈  135  weigh more   Apr 27, 2022
#2
+4

2. Body mass index, or BMI, of 18-year-old females are normally distributed with a mean of 24.6 and
a standard deviation of 2.4. Eighteen-year-old Sasha BMI has a z-score of 1.22.
What is Sasha’s BMI?

[  Sash's BMI   -  24.6 ] /  2.4 =    1.22

Sasha's BMI  -24.6  =   1.22 ( 2.4)

Sasha's BMI =   1.22 (2.4)  + 24.6  ≈  27.53   Apr 27, 2022
#3
+3

3. An experiment compares weight of fish for two different brands fish food. Nacho Average Minos
fish food has a mean fish weight 12 lbs., and a standard deviation of 1.4 lbs. Impossible Minos fish
food has a mean fish weight of 14 lbs. and a standard deviation of 1.2 lb. A fish is measured to be
13.1 lbs. Which brand of fish food, Nacho Average Minos, or Impossible Minos, is more likely to have
been used to feed this fish?

Nacho Average Minos  =  [ 13.1  - 12 ] / 1.4 =  .785

Impossible Minos   =  [ 13.1  - 14 ]  /  1.2 =  - .75

Impossible Minos , because  the fish measured   falls closer to  the  average   Apr 27, 2022
#4
+2

4. Height of 10th grade boys is normally distributed with a mean of 66.9 in. and a standard deviation
of 2.5 in.
The area greater than the z-score is the probability that a randomly selected 15-year-old boy
exceeds 71 in.
What is the probability that a randomly selected 10th grade boy exceeds 71 in.?

[ 71  -  66.9  ] / 2.5  =  1.8  = z score

Table  vaue for this  z score  =  .8810

Probability  =  1  - .8810   =  .119  =   11.9%   Apr 27, 2022
#5
+2

Thank you so much CPhill!!!! I really appreciate it! #6
+3

5. Heights for 16-year-old girls are normally distributed with a mean of 64 in. and a standard
deviation of 1.9 in.
Find the z-score associated with the 90th percentile.
Find the height of a 16-year-old girl in the 90th perc

z score associated with  .90  = 1.29

( Height -  64 ) /  1.9   =  1.29

Height   -64  =   1.29 * 1.9

Height  =   1.29 ( 1.9)  + 64   ≈    66.4  in   Apr 27, 2022