Question

According to a study done by De Anza students, the height for
Asian adult males is normally distributed with an average of 66
inches and a standard deviation of 2.5 inches. Suppose one Asian
adult male is randomly chosen. Let *X* = height of the
individual.

1. Find the probability that the person is between 64 and 68
inches. (Round to four decimal places) and Sketch the Graph.

2. Would you expect to meet many Asian adult males over 72 inches? Explain why or why not, and justify your answer numerically.

No, because the probability that an Asian male is over 72 inches tall is_______

3. The middle 40% of heights fall between what two values?

Write the probability statement.

* P*(

x1=

x2=

Answer #1

X ~ N(66, 2.5)

1. We need to find: P(64 < X < 68)

When X = 64, z = (64 - 66)/2.5 = -0.8

When X = 68, z = (68 - 66)/2.5 = 0.8

Hence, P(-0.8 < z < 0.8)

= P(z < 0.8) - P(z < -0.8)

From the z-table,

**= 0.7881 - 0.2119
= 0.5762**

2. We need to find: P(X > 72)

When X = 72, z = (72 - 64)/2.5 = 3.2

Hence, P(z > 3.2)

= 1 - P(z < 3.2)

= 1 - 0.9993

= 0.0007

No, because the probability that an asian male is over 72 inches tall is 0.0007

3. We need to find:

P(z1 < z < z2) = 0.40

By symmetry of normal distribution,

P(0 < z < z2) = P(z1 < z < 0) = 0.20

P(z < z2) = 0.5 + 0.20 = 0.70

P(z < 0) - P(z < z1) = 0.20

P(z < z1) = 0.30

From z-table,

z2 = 0.525

z1 = -0.525

X2 = 67.3125

X1 = 64.6875

P(64.69 < X < 67.31)

According to a study done by De Anza students, the height for
Asian adult males is normally distributed with an average of 66
inches and a standard deviation of 2.5 inches. Suppose one Asian
adult male is randomly chosen. Let X = height of the
individual.
Give the distribution of X
X~ ______ (_____,_____)
Find the probability that the person is between 65 and 69
inches.
Write the probability statement.
P(_______<X<_________)
What is the probability (round to four decimals
places)...

According to a study done by De Anza students, the height for
Asian adult males is normally distributed with an average of 66
inches and a standard deviation of 2.5 inches. Suppose one Asian
adult male is randomly chosen. Let X = height of the
individual.
The middle 40% of heights fall between what two values?
Write the probability statement.
P(x1 < X < x2) =
State the two values. (Round your answers to one decimal
place.)
x1
=
x2...

According to a study done by UCB students, the height for
Martian adult males is normally distributed with an average of 65
inches and a standard deviation of 2.4 inches. Suppose one Martian
adult male is randomly chosen. Let X = height of the individual.
Round all answers to 4 decimal places where possible. a. What is
the distribution of X? X ~ N( , ) b. Find the probability that the
person is between 65.5 and 68.5 inches. c....

According to a study done by UCB students, the height for
Martian adult males is normally distributed with an average of 68
inches and a standard deviation of 2.5 inches. Suppose one Martian
adult male is randomly chosen. Let X = height of the individual.
Round all answers to 4 decimal places According to a study done by
UCB s% of Martian heights lie between what two numbers? Low:
__inches High: ___inches must be 4 decimals rounded

According to the National Health Survey, the heights of adult
males in the United States are normally distributed with mean 69.0
inches and a standard deviation of 2.8 inches.
(a) What is the probability that an adult male chosen at random
is between 66 and 72 inches tall? (Round your answer to five
decimal places.)
(b) What percentage of the adult male population is more than 6
feet tall? (Round your answer to three decimal places.)

An adult height in North America is normally distributed with a
mean of 65 inches and a standard deviation of 3.5 inches.
1. Find the probability that the height is between 64 and 66 inches
P(64<X<66)
2. Find the probability that the height is greater than 70 inches.
P(X>70).
3. The middle 40% of heights fall between what two values? P(x1
< X < x2)
so beyond lost on this one

A large study of the heights of 920 adult men found that the
mean height was 71 inches tall. The standard deviation was 7
inches. If the distribution of data was normal, what is the
probability that a randomly selected male from the study was
between 64 and 92 inches tall? Use the 68-95-99.7 rule (sometimes
called the Empirical rule or the Standard Deviation rule). For
example, enter 0.68, NOT 68 or 68%.

To estimate the mean height μμ of male students on your campus,
you will measure an SRS of students. Heights of people of the same
sex and similar ages are close to Normal. You know from government
data that the standard deviation of the heights of young men is
about 2.8 inches. Suppose that (unknown to you) the mean height of
all male students is 70 inches.
(a) If you choose one student at random, what is the probability
that...

Please refer to this scenario in order to answer all questions
below: Given that the heights of males ages 18-25 (measured in
inches) have the distribution N(69.5, 1.5), find each of the
following. This notation means that it follows a normal curve,
their mean height is 69.5 inches, and their standard deviation is
1.5 inches. Draw a normal curve first and use that to answer the
questions. Make sure that you label the x-axis correctly. You will
need to find...

The average height of an adult male in the United States is 70
inches, with a
standard deviation of 3 inches. Assume that male heights are
Normally distributed.
Approximately what proportion of males are expected to be under
76 inches? To aid your answer, hand-draw to sketch a Normal
curve, and shade in the area under the Normal density curve the
question represents. Add dashed lines at the mean +/- 1SD, 2SD and
3SD. Then calculate the proportion asked about...

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 24 minutes ago

asked 24 minutes ago

asked 44 minutes ago

asked 52 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 2 hours ago

asked 2 hours ago

asked 2 hours ago

asked 2 hours ago

asked 2 hours ago