Answer:
The concentration of salt in the tank approaches
Step-by-step explanation:
Data provide in the question:
Water contained in the tank = 8000 L
Salt per litre contained in Brine = 35 g/L
Rate of pumping water into the tank = 25 L/min
Concentration of salt
Now,
Dividing both numerator and denominator by , we have
Here,
The concentration of salt in the tank approaches
Using the normal distribution and the central limit theorem, it is found that there is a 0.0284 = 2.84% probability of finding a sample mean mass of 695g or below.
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Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem establishes 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.
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The probability of finding a sample mean mass of 695g or below is the p-value of Z when X = 695, thus:
By the Central Limit Theorem
has a p-value of 0.0284.
0.0284 = 2.84% probability of finding a sample mean mass of 695g or below.
A similar problem is given at brainly.com/question/22934264
Answer:
(a)
(b)
(c)
(d)
Step-by-step explanation:
Given
--- at least
Solving (a): 28 combinations
From the question, we understand that; a combination of 28 is to be selected. Because the order is not important, we make use of combination.
Also, because repetition is allowed; different balloons of the same kind can be selected over and over again.
So:
Solving (b): At most 12 red balloons
First, we calculate the ways of selecting at least 13 balloons
Out of the 28 balloons, there are 15 balloons remaining (i.e. 28 - 13)
So:
Selection of at least 13 =
Ways of selecting at most 12 =
--- Complement rule
Solving (c): At most 8 blue balloons
First, we calculate the ways of selecting at least 9 balloons
Out of the 28 balloons, there are 19 balloons remaining (i.e. 28 - 9)
So:
Selection of at least 9 =
Ways of selecting at most 8 =
--- Complement rule
Solving (d): 12 red and 8 blue balloons
First, we calculate the ways for selecting 13 red balloons and 9 blue balloons
Out of the 28 balloons, there are 6 balloons remaining (i.e. 28 - 13 - 9)
So:
Selection =
Using inclusion/exclusion rule of two sets: