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AlekseyPX
3 years ago
10

The volume of wine in liters produced by a parcel of vineyard every year is modeled by a Gaussian distribution with an average o

f 100 and a variance of 9. Find the probability that this year it will produce 115 liters of wine
Mathematics
1 answer:
saul85 [17]3 years ago
6 0

Answer:

0.99865

Step-by-step explanation:

The question above is modelled by gaussian distribution. Gaussian distribution is also known as Normal distribution.

To solve the above question, we would be using the z score formula

The formula for calculating a z-score

z = (x-μ)/σ,

where x is the raw score

μ is the population mean

σ is the population standard deviation.

In the above question,

x is 115 liters

μ is 100

σ is the population standard deviation is unknown. But we were given variance in the question.

Standard deviation = √Variance

Variance = 9

Hence, Standard deviation = √9 = 3

We go ahead to calculate our z score

z = (x-μ)/σ

z = (115 - 100) / 3

z = 15/ 3

z score = 5

Using the z score table of normal distribution to find the Probability of having a z score of 5

P(x = 115) = P(z = 5) =

0.99865

Therefore the probability that this year it will produce 115 liters of wine = 0.99865

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Answer:

Step-by-step explanation:

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BigorU [14]

Answer:

(a) P (none) = 0.005873

(b) P ( at least one household ) = 0.9941

(c) P ( at most one household )  = 0.9960

(d) not wrong

Step-by-step explanation:

Given data

probability P = 29%  = 0.29

no of sample n = 15

tv show  X = 50

to find out

the probability that none of the households

the probability that at least one household

the probability that at most one household

If at most one household is tuned to 50 Minutes, does it appear that the 19% share value is wrong

solution

the probability that none of the households is tuned to 50 Minutes is

P (none) = 10C0 × P^{0}  ×  (1 - 0.29)^{15}

P (none) =  (0.71)^{15}

P (none) = 0.005873

and the probability that at least one household is tuned to 50 Minutes is

P ( at least one household ) = 1 - P(none)

P ( at least one household ) = 1 - 0.005873

P ( at least one household ) = 1 - 0.005873

P ( at least one household ) = 0.9941

the probability that at most one household is tuned to 50 Minutes is

P ( at most one household ) = P (none)  + P (x=1)

P ( at most one household )  =0.9941 + 10C1 × (0.23)^{1} × (0.71)^{14}

P ( at most one household )  = 0.9960

and in last part If at most one household is tuned to 50 Minutes it appear that the 19% share value is not wrong

because P( at most one household ) is not zero

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