What is the surface area of the right cone with the a raidus of 3 and a slant hieght of 15
1 answer:
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Answer:
15.15% probability that the sample mean will be $192,000 or more.
Step-by-step explanation:
To solve this problem, 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 random variable X, with mean and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
In this problem, we have that:
The probability that the sample mean will be $192,000 or more is
This is 1 subtracted by the pvalue of z when X = 192000. So
By the Central Limit Theorem
has a pvalue of 0.8485.
1-0.8485 = 0.1515
15.15% probability that the sample mean will be $192,000 or more.
Answer:
Step-by-step explanation:
The given parameters can be represented as:
Solving (a): P(x < 3)
This is calculated as:
----- i.e. all probabilities less than 3
So, we have:
Solving (b): Expected number of events
This is calculated as:
So, we have:
Express as decimal
Approximate to the nearest integer
Answer:
The probability will be 0.3085 or 0
Step-by-step explanation:
Given:
True mean=12.5
Sample mean =12.6
Standard deviation=0.2
Samples=100
To Find:
Probability that exceeds 12.6 ounces.
Solution:
Calculate the Z-score for given means and standard deviation.
So
Z-score= (true mean -sample mean)/standard deviation.
Z-score=(12.5 -12.6)/0.2
=-0.1/0.2
=-0.5
Now Using Z-table
P(X≥-0.5)=p(Z≥-0.5)=0.3085
Hence Probability that sample mean weight exceeds will be 0.3085
OR
By using Normal distribution with sampling ,it will be as follows
Z=(X-u)/[Standard deviation/Sqrt(No of samples)]
Z=(12.6-12.5)/(0.2/Sqrt(100)
Z=0.1/0.2/10
Z=5
So P(X≥12.6 )=P(Z≥5)=1
Pr(Z≥5)=1-1=0.
(Refer the attachment )
Hence Probability of getting ounces greater than 12.6 is '0'.
The sampling is of 0.02 size hence graphically it looks likely.
as shown in attachment.
