Question

The ages of all college students follow a normal distribution with a mean 26 years and a standard deviation of 4 years. Find the probability that the mean age for a random sample of 36 students would be between 25 and 27.

Answer #1

The ages of all college students follow a normal distribution
with a mean of 27 years and standard deviation of 4.4 years. Find
the probability that the mean age for a random sample of 36
students would be:
Between 25.5 and 28 years old
Less than 25.5 years old
Instructions:
Draw the normal curve and shade the area where applicable
Do not give calculator key strokes. (Give the formula used)
Circle your answer to the question.

Suppose students' ages follow a skewed right distribution with a
mean of 24 years old and a standard deviation of 5 years. If we
randomly sample 250 students, which of the following statements
about the sampling distribution of the sample mean age is
incorrect?
The shape of the sampling distribution is approximately
normal.
The mean of the sampling distribution is approximately 24 years
old.
The standard deviation of the sampling distribution is equal to
5 years.
All of the above...

Suppose students ageas follow a skewed right distribution with a
mean of 25 years old and a standard deviation of 15 years. consider
the random sample of 100 students.
Determine the probability that the sample mean student age is
greater than 22 years?

Listed below are the ages of Dortmund College Students
27,31,19,21,26,25,20,18,17,45,23,26,21,20,19,25,24,26,51,39,17
Assuming there is a normal distribution for these ages, find the
probability that a student is at least 25 years old.

A college admissions director wishes to estimate the mean age of
all students currently enrolled. In a random sample of 19 students,
the mean age is found to be 23.7 years. From past studies, the ages
of enrolled students are normally distributed with a standard
deviation of 10.4 years. Construct a 90% confidence interval for
the mean age of all students currently enrolled.
1. The critical value:
2. The standard deviation of the sample mean:
3. The margin of error...

Ages of students: A simple random sample of 100
U.S. college students had a mean age of 22.68 years. Assume the
population standard deviation is σ = 4.74 years. Construct
a 99% confidence interval for the mean age of U.S. college
students.
The answer I posted my instructor says it is wrong. I came up
with 21.45 to 23.91 a 99% confidence interval for the mean of the
U.S college students
She said the z procedures are needed.
Thanks.

A college admissions director wishes to estimate the mean age of
all students currently enrolled. In a random sample of 22 students,
the mean age is found to be 21.4 years. From past studies, the ages
of enrolled students are normally distributed with a standard
deviation of 10.2 years. Construct a 90% confidence interval for
the mean age of all students currently enrolled.
1. What is the Critical value?

6. Almost all medical schools in the United States require
students to take the Medical College Admission Test (MCAT). The
scores follow a normal distribution. From published information the
standard deviation is 6.4. On a particular campus, the mean score
of those taking the MCAT is 25.0.
a. If a student is chosen at random, what is the probability the
individual’s score is between 20 and 30?
b. In a sample of 25 students, what is the sampling distribution
of...

The scores of fourth grade students on a mathematics achievement
test follow a normal distribution with a mean of 75 and standard
deviation of 4. What is the probability the sample mean score of 64
randomly selected students is between 74 and 76?

The credit card balance for college students has a normal
distribution with a mean of $3450 and a standard deviation of $995.
A random sample of 37 students is selected.
1）Describe the sampling distribution of the sample mean for the
37 students
A. X ~ AN(3450,995)
B. X ~ N(3450,163.577)
C. X ~ AN(3450,163.577)
D. X ~ N(3450,995)
2）Find the probability that the mean credit card balance for the
sample of 37 students is more than $3100.
A. 0.6368
B....

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