two numbers rational. between 0.3 and 0.4 express one as a decimal and fraction. explain why these numbers are rational
1 answer:
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<h2>Answer:</h2>
The different possible routes are :
181440
<h2>Step-by-step explanation:</h2>We are asked to find the number of different routes that are possible to choose 7 locations among the 9 locations remaining.
This means that we have to chose the locations and also we have to arrange them in different orders according to the different order of locations.
The method we have to use is a method of permutation.
When we have to chose and arrange r items out of a total of n items then the formula is given by:
Here we have: n=9
and r=7
Hence, the different possible routes are:
Hence, the answer is:
181440
Answer:
a) 0.0016 = 0.16% probability that the sample mean is at least $30.00.
b) 0.8794 = 87.94% probability that the sample mean is greater than $26.50 but less than $30.00
c) 90% of sample means will occur between $26.1 and $28.9.
Step-by-step explanation:
To solve this question, we need to understand 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.
In this question, we have that:
a. What is the likelihood the sample mean is at least $30.00?
This is 1 subtracted by the pvalue of Z when X = 30. So
By the Central Limit Theorem, we have that:
has a pvalue of 0.9984
1 - 0.9984 = 0.0016
0.0016 = 0.16% probability that the sample mean is at least $30.00.
b. What is the likelihood the sample mean is greater than $26.50 but less than $30.00?
This is the pvalue of Z when X = 30 subtracted by the pvalue of Z when X = 26.50. So
From a, when X = 30, Z has a pvalue of 0.9984
When X = 26.5
has a pvalue of 0.1190
0.9984 - 0.1190 = 0.8794
0.8794 = 87.94% probability that the sample mean is greater than $26.50 but less than $30.00.
c. Within what limits will 90 percent of the sample means occur?
Between the 50 - (90/2) = 5th percentile and the 50 + (90/2) = 95th percentile, that is, Z between -1.645 and Z = 1.645
Lower bound:
Upper Bound:
90% of sample means will occur between $26.1 and $28.9.