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qaws [65]
4 years ago
7

A bottling company uses a filling machine to fill plastic bottles with cola. The bottles are supposed to contain 300 milliliters

(ml). In fact, the contents vary according to a Normal distribution with mean μ = 298 ml and standard deviation σ = 3 ml. What is the probability that the mean contents of six randomly selected bottles is less than 295 ml? a. 0.0478 b. 0.9928 c. 0.1587 d. 0.00002 e. 0.0072
Mathematics
1 answer:
PilotLPTM [1.2K]4 years ago
7 0

Answer:

e. 0.0072

Step-by-step explanation:

We are given that a bottling company uses a filling machine to fill plastic bottles with cola. And the contents vary according to a Normal distribution with Mean, μ = 298 ml and Standard deviation, σ = 3 ml .

 Let    Z = \frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)  where, Xbar = mean contents of six randomly

                                                                     selected bottles

                                                          n = sample size i.e. 6    

So, Probability that the mean contents of six randomly selected bottles is less than 295 ml is given by, P(Xbar < 295)

 P(Xbar < 295) = P( \frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } } < \frac{295 - 298}{\frac{3}{\sqrt{6} } } ) = P(Z < -2.45) = P(Z > 2.45)

Now, using z% score table we find that P(Z > 2.45) = 0.00715 ≈ 0.0072 .

Therefore, option e is correct .

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Answer: f(n) = 0.9 × f(n − 1) + 10,   f(0) = 175,  n > 0


Step-by-step explanation:

Given: A store had 175 cell phones in the month of January.

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Let n be the number of months

Then when n=0

f(0)=175

After first month , n=1

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=f(0)-0.1\timesf(0)+10\\=f(0)(1-0.1)+10\\=f(0)(0.9)+10\\

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f(n)=0.9×f(n−1)+10, f(0) = 175, n > 0

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