Danica has laid out floor tiles so they form a rectangle with a perimeter of 18 inches.what is the differnce between the greates
2 answers:
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A="b is in the middle"
B="c is to the right of b"
C="The letter def occur together in that order"
a) b can be in 7 places, but only one is the middle. So, P(A)=1/7
b) X=i, "b is in the i-th position"
Y=j, "c is in the j-th position"
P(B)=1/2
c) X=i, "d is in the i-th position"
Y=j, "e is in the j-th position"
Z=k, "f is in the i-th position"
P(C)=1/42
P(A∩C)=2*(1/7*1/6*1/5*1/4)=1/420
P(B∩A)=3*(1/7*1/6)=1/14
P(A|C)=P(A∩C)/P(C)=(1/420)/(1/42)=1/10
P(B|C)=P(B∩C)/P(C)=(1/420)/(1/42)=1/10
P(A|B)=P(B∩A)/P(B)=(1/14)/(1/2)=1/7
P(A∩B)=1/14
P(A)P(B)=(1/7)*(1/2)=1/14
A and B are independent
P(A∩C)=1/420
P(A)P(C)=(1/7)*(1/42)=1/294
A and C aren't independent
P(B∩C)=1/420
P(B)P(C)=(1/2)*(1/42)=1/84
B and C aren't independent
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%.