Part 2 of 2 (b) Find the 70" percentile of the number of apps. Round the answer to two decimal places. The 70" percentile of the number of apps is
Part 2 of 2 (b) Find the 70" percentile of the number of apps. Round the answer to two decimal places. The 70" percentile of the number of apps is
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
![How many apps? According to a website, the mean number of apps on a smartphone in the United States is 90. Assume the number of apps is normally
distributed with mean 90 and standard deviation 21.
6.1 The Normal Curve
Convert nermativalues tojssCAresii bri7 oT
Solve applications using the normal curve
100
rotal area
0.35
0.30-
0.25-
0.20-
0.15-
Probebility
0.10-
0.05
Area=0.16
91032-S 16 ripiA entot be9TA erli bniloT
10
a. What proportion of the population is between 4 and 6?
O. 16
67 P(4Lĕ E6) = 0.16 =16 7.
c. What proportion of the population is not between 4 and 6?
1-0.1b =0 84
owT neewte8 691A edt þai7 OT
6?
0.84
Tx=
The Normal Distribution
bell shaper
Symmearic
for POPUlationg that
2-Score i Ħ Of
Steindaro devaion
from the
Used
mean
A the distribution of a continuous variable.
Shade area under the curve and use Table A.2 to find the area/probability
b. If a value is at is the that it will be b
distributed](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1efb7724-8be7-4468-9630-253c0ee9f8b8%2F77fff921-48a6-457b-bb29-d37b1f609a8c%2Fwl07pcf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:How many apps? According to a website, the mean number of apps on a smartphone in the United States is 90. Assume the number of apps is normally
distributed with mean 90 and standard deviation 21.
6.1 The Normal Curve
Convert nermativalues tojssCAresii bri7 oT
Solve applications using the normal curve
100
rotal area
0.35
0.30-
0.25-
0.20-
0.15-
Probebility
0.10-
0.05
Area=0.16
91032-S 16 ripiA entot be9TA erli bniloT
10
a. What proportion of the population is between 4 and 6?
O. 16
67 P(4Lĕ E6) = 0.16 =16 7.
c. What proportion of the population is not between 4 and 6?
1-0.1b =0 84
owT neewte8 691A edt þai7 OT
6?
0.84
Tx=
The Normal Distribution
bell shaper
Symmearic
for POPUlationg that
2-Score i Ħ Of
Steindaro devaion
from the
Used
mean
A the distribution of a continuous variable.
Shade area under the curve and use Table A.2 to find the area/probability
b. If a value is at is the that it will be b
distributed
![Part 2 of 2
(b) Find the 70™ percentile of the number of apps. Round the answer to two decimal places.
The 70th percentile of the number of apps is.
Shade area unde
Convert normal value
Solve applications using
6.1 The N
Learning Objective
0.35-
Probability
0.30-
0.25-
0.20-
0.15-
Area=0.16
0.10-
0.05-
0-
What proportion of the population is between 4 and 67
0.16
b. If a value is chosen at random from this population, what is
62
P(4LX 56)=0.16 - 16%
AD
c. What proportion of the population is not between 4 and67
91'0-1
3D
084
d. If a value is chosen at random from this population, what
62
0.341
The Normal Distribution
147%
bell Shape
Simmentric
eo for PopUloーかns tr
2-Suore E
Sendord des
A probability density curve represents the prol
Fry](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1efb7724-8be7-4468-9630-253c0ee9f8b8%2F77fff921-48a6-457b-bb29-d37b1f609a8c%2Fmb5tyz_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Part 2 of 2
(b) Find the 70™ percentile of the number of apps. Round the answer to two decimal places.
The 70th percentile of the number of apps is.
Shade area unde
Convert normal value
Solve applications using
6.1 The N
Learning Objective
0.35-
Probability
0.30-
0.25-
0.20-
0.15-
Area=0.16
0.10-
0.05-
0-
What proportion of the population is between 4 and 67
0.16
b. If a value is chosen at random from this population, what is
62
P(4LX 56)=0.16 - 16%
AD
c. What proportion of the population is not between 4 and67
91'0-1
3D
084
d. If a value is chosen at random from this population, what
62
0.341
The Normal Distribution
147%
bell Shape
Simmentric
eo for PopUloーかns tr
2-Suore E
Sendord des
A probability density curve represents the prol
Fry
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