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solmaris [256]
3 years ago
11

The volume of soda a dispensing machine pours into a 12-ounce can of soda follows a normal distribution with a mean of 12.45 oun

ces and a standard deviation of 0.30 ounce. The company receives complaints from consumers who actually measure the amount of soda in the cans and claim that the volume is less than the advertised 12 ounces. What proportion of the soda cans contain less than the advertised 12 ounces of soda?
Mathematics
1 answer:
lorasvet [3.4K]3 years ago
6 0

Answer: 0.0668

Step-by-step explanation:

Given : The volume of soda a dispensing machine pours into a 12-ounce can of soda follows a normal distribution with a mean of 12.45 ounces and a standard deviation of 0.30 ounce.

i.e. \mu=12.45 and \sigma=0.30

Let x denotes the volume of soda a dispensing machine pours into a 12-ounce can.

Then, the proportion of the soda cans contain less than the advertised 12 ounces of soda will be :-

P(x

Hence, the proportion of the soda cans contain less than the advertised 12 ounces of soda = 0.0668

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In a particular faculty 60% of students are men and 40% are women. In a random sample of 50 students what is the probability tha
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a) The expected value is given by:

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and the variance is given by:

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c) 1) Random sample (assumed)

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Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

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X \sim Binom(n=50, p=0.4)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

Part a

The expected value is given by:

E(X) = np = 50*0.4 = 20

and the variance is given by:

Var(X) =np(1-p) = 50*0.4*(1-0.4) = 12

Part b

For this case we want to find this probability:

P(X>25)= 1-P(X\leq 25)

And we can find this probability with the following Excel code:

=1-BINOM.DIST(25,50,0.4,TRUE)

And we got:

P(X>25)= 1-P(X\leq 25)=0.0573

Part c

1) Random sample (assumed)

2) np= 50*0.4= 20 >10

n(1-p) =50*0.6= 30>10

3) Independence (assumed)

Since the 3 conditions are satisfied we can use the normal approximation:

X \sim N(\mu = 20 , \sigma= 3.464)

Part d

We want this probability:

P(X>25) = 1-P(Z< \frac{25-20}{3.464}) = 1-P(z

Part e

For this case we use the continuity correction and we have this:

P(X>25)= P(X>25.5) = 1-P(X \leq 25.5)

P(X>25)= P(X>25.5) = 1-P(X \leq 25.5)= 1-P(Z

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