(08.06)The following data show the height, in inches, of 11 different garden gnomes:
1 answer:
Answer:
The correct statement is:
On average, the height of a garden gnome varies 3.2 inches from the mean of 6 inches.
Step-by-step explanation:
We are given a data of 11 gardens as:
2 9 1 23 3 7 10 2 10 9 7
Now on removing the outlier i.e. 23 (since it is the very large value as compared to other data points) the entries are as follows:
x |x-x'|
2 4
9 3
1 5
3 3
7 1
10 4
2 4
10 4
9 3
7 1
Now mean of the data is denoted by x' and is calculated as:
Hence, Mean(x')=6
Now,
∑ |x-x'|=32
Now mean of the absolute deviation is:
This means that , On average, the height of a garden gnome varies 3.2 inches from the mean of 6 inches.
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Answer:
To use a Normal distribution to approximate the chance the sum total will be between 3000 and 4000 (inclusive), we use the area from a lower bound of 3000 to an upper bound of 4000 under a Normal curve with its center (average) at 3500 and a spread (standard deviation) of 160 . The estimated probability is 99.82%.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are solved using 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.
For sums, we have that the mean is and the standard deviation is
In this problem, we have that:
This probability is the pvalue of Z when X = 4000 subtracted by the pvalue of Z when X = 3000.
X = 4000
By the Central Limit Theorem
has a pvalue of 0.9991
X = 3000
has a pvalue of 0.0009
0.9991 - 0.0009 = 0.9982
So the correct answer is:
To use a Normal distribution to approximate the chance the sum total will be between 3000 and 4000 (inclusive), we use the area from a lower bound of 3000 to an upper bound of 4000 under a Normal curve with its center (average) at 3500 and a spread (standard deviation) of 160 . The estimated probability is 99.82%.
Answer:
h=90