There are 40 dimes and quarters in a drawer. The total amount of the coins
1 answer:
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If patio is square it means that area of if is 
where a is one side
Now we can substitute A=200 and solve eq
a=
a=14.1421
Now we are rounding to the nearest tenths
a=14.1 because next number is 4 and is less then 5.
The result is 14.1
Now we can substitute A=200 and solve eq
a=
a=14.1421
Now we are rounding to the nearest tenths
a=14.1 because next number is 4 and is less then 5.
The result is 14.1
Answer:
11.51% probability that the mean annual salary of the sample is between $71000 and $73500
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 question, we have that:
What is the probability that the mean annual salary of the sample is between $71000 and $73500?
This is the pvalue of Z when X = 73500 subtracted by the pvalue of Z when X = 71000. So
X = 73500
By the Central Limit Theorem
has a pvalue of 0.1151
X = 71000
has a pvalue of 0.
0.1151 - 0 = 0.1151
11.51% probability that the mean annual salary of the sample is between $71000 and $73500