Frank, Carl, and Vinny started a lawn mowing business over the summer. This table shows how many lawns each worker mowed during
the summer months.
Display the data in matrix W with rows indicating workers. What is element w13?
a.22
b.29
c.21
d.19
Display the data in matrix W with rows indicating workers. What is element w13?
a.22
b.29
c.21
d.19
2 answers:
Answer:
d. 19
Step-by-step explanation:
A matrix is an array of rectangular shape which consists of numbers, symbols, or expressions, arranged in rows and columns.
means element in first row third column. First row means Frank and third column means August.
So, the number of lawns mowed by Frank in August is
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Answer:
0.0174 = 1.74% probability that the sample mean would be greater than 101.63 WPM
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.
In this problem, we have that:
What is the probability that the sample mean would be greater than 101.63 WPM?
This is 1 subtracted by the pvalue of Z when X = 101.63. So
By the Central Limit Theorem
has a pvalue of 0.9826.
1 - 0.9826 = 0.0174
0.0174 = 1.74% probability that the sample mean would be greater than 101.63 WPM