SAT College

Find the range and standard deviation.

The table below gives the number of hours spent watching TV last week by a sample of 24 children.

9 5 6 7 2 4
8 6 6 4 5 1
4 8 3 2 10 7
8 9 2 9 7 7


Range =
(Please enter an exact answer.)

Standard Deviation =
(Please show your answer to 4 decimal places.)

Answers

Answer 1

The range of the data is 9.

The standard deviation is 3.1300

What is Standard Deviation?

Quantifying the amount of variability or dispersion within data sets is achieved through a statistical tool known as standard deviation.

This popular method calculates how distant individual data points are from the mean or average value, signifying whether they are closely congregated or spread out over an extensive interval.

Notably, when the data has a small standard deviation, it means that the values lie within proximity to the mean.

Conversely, a higher value indicates a wide variance between the different numbers in the set.

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Find The Range And Standard Deviation.The Table Below Gives The Number Of Hours Spent Watching TV Last

Related Questions

What is permissive mean

Answers

The term "permissive" generally means granting permission or allowing something to be done without objection or interference. It implies a degree of flexibility or leniency in terms of rules or regulations. For example, a permissive parent may allow their child to make their own decisions without strict rules or discipline. In contrast, an authoritarian parent may have strict rules and expect strict obedience from their child. In a legal context, a permissive law or regulation allows something to be done, but does not necessarily require it.

A large collection of one-digit random numbers should have about 50% odd and 50% even digits because five of the ten digits are odd (1, 3, 5, 7, and 9) and five are even (0, 2, 4, 6, and 8).
a. Find the proportion of odd-numbered digits in the following lines from a random number table. Count carefully.
8 2 6 6 9
0 9 2 2 4
9 3 8 6 8
4 3 9 6 3
2 6.9 3 6
1 5 8 1 6

The given random number table consists of ___% odd-numbered digits (Round to two decimal places as needed.)

Does the proportion found in part (a) represent ^p (the sample proportion) or p (the population proportion)?

Find the error in this estimate, the difference between ^p and p (or ^p - p).

Answers

The given random number table consists of 50 % odd-numbered digits, which represents the population proportion because it is based on the entire population of one-digit random numbers.

There is no error in this estimate since the sample proportion (^p) is equal to the population proportion (p) in this case.

What is the explanation for the aforegoing?

 (a) Counting the odd  -numbered digits  in each line, we get:

2/5, 2 /5,  3 /5, 3/ 5, 2/5,   2 /5

Therefore, the proportion of odd - numbered digits in the given random number table is:

(2/5 + 2/5 + 3/ 5 + 3/5 + 2/5 + 2/5) / 6 = 14/30 =  46.67 %

(b) The proportion found in part (a) represents ^p (the sample proportion).

(c) The population proportion is p = 0.5 (since we expect 50% odd and 50% even digits in the population). The error in the estimate is the difference between ^p and p:

^p - p = 0.467 - 0.5

=  -3.3%

Thus,  there is no error in this estimate since the sample proportion (^p) is equal to the population proportion (p) in this case.

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According to the Bureau of Labor Statistics, 71.9% of Young women enroll in college directly after high school graduation. Suppose a random sample of 200 female high school graduates is selected and the proportion who enroll in college is obtained.


a. What value should we expect for the sample proportion?

b. What is the standard error?

c. What effect would increasing the sample size to 500 have on the standard error?

Answers

a. The expected value for the sample proportion can be calculated as the population proportion: 71.9%, so we expect that approximately 71.9% of the 200 female high school graduates will enroll in college directly after high school graduation.

Expected value = population proportion * sample size = 0.719 * 200 = 143.8

Therefore, we expect approximately 144 female high school graduates to enroll in college directly after high school graduation.

b. The standard error of the sample proportion can be calculated as:

SE = sqrt[p(1-p)/n], where p is the population proportion and n is the sample size.

SE = sqrt[0.719(1-0.719)/200] = 0.033

Therefore, the standard error is 0.033.

c. The standard error is inversely proportional to the square root of the sample size. Increasing the sample size from 200 to 500 would reduce the standard error:

SE = sqrt[0.719(1-0.719)/500] = 0.023

Therefore, the standard error would decrease to 0.023 if the sample size increased to 500.
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