Which number is a common factor of 24, 48, and 144? 16 10 9 8
1 answer:
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Answer:
83.89% probability that the sample mean would be greater than 98.8 gallons
Step-by-step explanation:
To solve this question, we have 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 random variable X, with mean and standard deviation
, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation, which is also called standard error
In this problem, we have that:
The standard deviation is the square root of the variance. So
If 44 racing cars are randomly selected, what is the probability that the sample mean would be greater than 98.8 gallons
This probability is 1 subtracted by the pvalue of Z when X = 98.8. So
By the Central Limit Theorem
has a pvalue of 0.1611
1 - 0.1611 = 0.8389
83.89% probability that the sample mean would be greater than 98.8 gallons
Answer:
(a) <u>Sampling distribution </u>
P(25) = 0,04
P(35) = 0.1 + 0.1 = 0,2
P(42,5) = 0.06 + 0.06 = 0,12
P(45) = 0,25
P(52,5) = 0.15 + 0.15 = 0,3
P(60) = 0,09
(b) E(X) = 45.5 oz
(c) E(X) = μ
Step-by-step explanation:
The variable we want to compute is
For this we need to know all the possible combinations of X1 and X2 and the probability associated with them.
(a) <u>Sampling distribution </u>
Calculating all the 9 combinations (3 repeated, so we end up with 6 unique combinations):
P(25) = P(X1=25) * P(X2=25) = p25*p25 = 0.2 * 0.2 = 0,04
P(35) = p25*p45+p45*p25 = 0.2*0.5 + 0.5*0.2 = 0.1 + 0.1 = 0,2
P(42,5) = p25*p60 + p60*p25 = 0.2*0.3 + 0.3*0.2 = 0.06 + 0.06 = 0,12
P(45) = p45*p45 = 0.5 * 0.5 = 0,25
P(52,5) = p45*p60 + p60*p45 = 0.5*0.3 + 0.3*0.5 = 0.15 + 0.15 = 0,3
P(60) = p60*p60 = 0.3*0.3 = 0,09
(b) Using the sample distribution, E(X) can be expressed as:
The value of E(X) is 45.5 oz.
(c) The value of μ can be calculated as
We can conclude that E(X)=μ
We could have arrived to this conclusion by applying