A line has a rise of 3 and a run of 12. find the slope of the line. type a numerical answer in the space provided. if necessary,
1 answer:
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Answer:
a) A sample of 256 was used in this survey.
b) 45.14% probability that the point estimate was within ±15 of the population mean
Step-by-step explanation:
This question is solved using the normal probability distribution and the central limit theorem.
Normal probability distribution
When the distribution is normal, we use the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
a. How large was the sample used in this survey?
We have that . We want to find n, so:
A sample of 256 was used in this survey.
b. What is the probability that the point estimate was within ±15 of the population mean?
15 is the bounds with want, 25 is the standard error. So
Z = 15/25 = 0.6 has a pvalue of 0.7257
Z = -15/25 = -0.6 has a pvalue of 0.2743
0.7257 - 0.2743 = 0.4514
45.14% probability that the point estimate was within ±15 of the population mean
Answer:
1. The shape of the histogram created with MS Excel is approximately bell shaped and approximately evenly spread about the central (largest counts) values
The shape of the box plot created with MS Excel is approximately evenly distributed
2. The appropriate measure of central tendency for a bell shaped histogram is the mean and median, due to the approximately equal distribution about the highest frequency class
3. Part A
The 15 numbers between 1 and 20 generated by a random generator are;
1, 8, 8, 15, 4, 18, 11, 6, 17, 1, 18, 15, 10, 12, 11
The measure of center is the mean = (1+8+8+15+4+18+11+6+17+1+18+15+10+12+11)/15 = 155/15 = 31/3
The measure of spread used is the variance, s² = 32.38
= 453.333
n = 15
s² = 453./(15 - 1) = 32.38
Part B
The measure of central tendency are the mean and the median because the size of the data (n = 15), is 75% of the population (N = 20) nd therefore the data is approximately normal and can be represented by the mean and the standard deviation
Step-by-step explanation:
