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.
Read more about range here:
https://brainly.com/question/2264373
#SPJ1
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).
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.
Learn more bout sample proportion:
https://brainly.com/question/6211716?
#SPJ1
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?
What is permissive mean