Question

A survey among freshmen at a certain university revealed that the number of hours spent studying the week before final exams was normally distributed with mean 40 and standard deviation 7 . Find percentile corresponding to p = 62 % of the number of hours studying. Write only a number as your answer. Round to two decimal places (for example: 20.81).

Answer #1

Let X denotes the number of hours spent studying the week before final exams. Then X is a normally distributed random variable with mean 40 and standard deviation 7.

Let X_{0.62} denotes the percentile corresponding to p =
62% of the number of hours studying. It indicates that 62% of the
number of hours spent studying the week before final exams will be
lesser than X_{0.62.} Representing this as probability,

P(X < X_{0.62}) = 0.62

Using the standard normal transformation,

Apply this to our case.

From the standard normal distribution table,

P(Z < 0.305481) = 0.62

Now equate the similar terms of the equations and solve.

X_{0.62} = 0.305481(7) + 40 = 42.14

The percentile corresponding to p = 62% of the number of hours
studying is **42.14.**

A survey among freshmen at a certain university revealed that
the number of hours spent studying the week before final exams was
normally distributed with mean 25 and standard deviation 15. A
sample of 36 students was selected. What is the probability that
the average time spent studying for the sample was between 29.0 and
30 hours studying? Round to 4 decimal places.

A survey among freshmen at a certain university revealed that
the number of hours spent studying the week before final exams was
approximately normally distributed with mean 25 and standard
deviation 7.
What proportion of students studied more than 20 hours?
What proportion of students studied less than 16 hours?
What proportion of the students studied between 25 and 40
hours?

a survey among freshmen at a certain university revealed that
the number of hours spent studying the week before finals exams was
normally distributed with mean 25 and standard deviation 15. a
sample of 36 students was selected. what is the probability that
the average time spent studying for the sample was between 28.0 and
30 hours studying? round to four decimal places.

A survey among freshmen at a certain university revealed that
the number of hours spent studying the week before final exams was
approximately normally distributed with mean 25 and standard
deviation 7.
a. What proportion of students studied more than 20 hours?
b. What proportion of students studied less than 16 hours?
c. What proportion of the students studied between 25 and 40
hours?

A survey among freshmen at a large university showed that the
number of hours spent studying the week before final exams was
normally distributed with a mean of 25 hours and standard deviation
of 7 hours. What is the probability a randomly selected freshman
spends between 35 and 43 hours studying the week before final
exams?
a.
0.0713
b.
0.0565
c.
0.1271
d.
0.8729

A survey among freshmen at a large university showed that the
number of hours spent studying the week before final exams was
normally distributed with a mean of 25 hours and standard deviation
of 7 hours. What is the probability a randomly selected freshman
spends between 35 and 43 hours studying the week before final
exams?
a.
0.0565
b.
0.8729
c.
0.0713
d.
0.1271

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25
and standard deviation
5
. Use the Cumulative Normal Distribution Table to answer the
following.
(a) Find the
96th
percentile of the number of hours studying.
(b) Find the
31st
percentile of the number of hours studying.
(c) Between what two values are the middle
88%
of the...

6.1 question 18
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A
survey among freshman at a large university showed the number of
hours spent studying the week before finals was normally
distributed with a mean of 25 hours in a standard deviation of
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The weights of broilers (commercially raised chickens ) are
approximately normally distributed with mean 1387 grams and
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a number as your answer Round to 4 decimal places ( for example
0.0048 ). Do not write as a percentage .
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