Amy measured a house and made a scale drawing. The scale of the drawing was 5 millimeters : 2 meters. In the drawing, the dining
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Answer:
a) 0.0951
b) 0.8098
c) Between $24.75 and $27.25.
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:
(a)
What is the likelihood the sample mean is at least $27.00?
This is 1 subtracted by the pvalue of Z when X = 27. So
By the Central Limit Theorem
has a pvalue of 0.9049
1 - 0.9049 = 0.0951
(b)
What is the likelihood the sample mean is greater than $25.00 but less than $27.00?
This is the pvalue of Z when X = 27 subtracted by the pvalue of Z when X = 25. So
X = 27
has a pvalue of 0.9049
X = 25
has a pvalue of 0.0951
0.9049 - 0.0951 = 0.8098
c)Within what limits will 90 percent of the sample means occur?
50 - 90/2 = 5
50 + 90/2 = 95
Between the 5th and the 95th percentile.
5th percentile
X when Z has a pvalue of 0.05. So X when Z = -1.645
95th percentile
X when Z has a pvalue of 0.95. So X when Z = 1.645
Between $24.75 and $27.25.
Solution:
we have been asked to find the number of distinguishable permutations of the letters m, i, s, s, i, s, s, i, p, p, i.
Here we can see
m appears 1 time.
i appears 4 times.
S appears 4 times.
p appears 2 times.
Total number of letters are 11.
we will divide the permutation of total number of letters by the permutation of the number of each kind of letters.
The number of distinguishable permutations
Hence the number of distinguishable permutations