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kkurt [141]
3 years ago
5

In 3 billion repetitions of an experiment, a random event occurred in 500 million cases.

Mathematics
2 answers:
Gre4nikov [31]3 years ago
6 0
<h2>Answer:</h2>

The expected probability of the complement of the event is:

                      \dfrac{5}{6}

<h2>Step-by-step explanation:</h2>

We know that for any event A and the complement of the event i.e. A^c the sum of the probabilities of both the events is equal to 1.

i.e. if P denote the probability of an event then we have:

P(A)+P(A^c)=1

Here we have the probability of event A as:

                P(A)=\dfrac{1}{6}

Hence,

\dfrac{!}{6}+P(A^c)=1\\\\\\i.e.\\\\\\P(A^c)=1-\dfrac{1}{6}\\\\\\i.e.\\\\\\P(A^c)=\dfrac{6-1}{6}\\\\\\i.e.\\\\\\\\P(A^c)=\dfrac{5}{6}

           Hence, the answer is:

                \dfrac{5}{6}

Eddi Din [679]3 years ago
5 0
5/6 is your other percentage
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7 0
3 years ago
Read 2 more answers
Suppose a geyser has a mean time between irruption’s of 75 minutes. If the interval of time between the eruption is normally dis
lesya [120]

Answer:

(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.

(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.

(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.

(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size

(e) The population mean may be larger than 75 minutes between irruption.

Step-by-step explanation:

We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.

(a) Let X = <u><em>the interval of time between the eruption</em></u>

So, X ~ Normal(\mu=75, \sigma^{2} =20)

The z-score probability distribution for the normal distribution is given by;

                            Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)

 

    P(X > 84 min) = P( \frac{X-\mu}{\sigma} > \frac{84-75}{20} ) = P(Z > 0.45) = 1 - P(Z \leq 0.45)

                                                        = 1 - 0.6736 = <u>0.3264</u>

The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

(b) Let \bar X = <u><em>sample time intervals between the eruption</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

           n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P(\bar X > 84 min)

 

    P(\bar X > 84 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{84-75}{\frac{20}{\sqrt{13} } } ) = P(Z > 1.62) = 1 - P(Z \leq 1.62)

                                                        = 1 - 0.9474 = <u>0.0526</u>

The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.

(c) Let \bar X = <u><em>sample time intervals between the eruption</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

           n = sample of time intervals = 20

Now, the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is given by = P(\bar X > 84 min)

 

    P(\bar X > 84 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{84-75}{\frac{20}{\sqrt{20} } } ) = P(Z > 2.01) = 1 - P(Z \leq 2.01)

                                                        = 1 - 0.9778 = <u>0.0222</u>

The above probability is calculated by looking at the value of x = 2.01 in the z table which has an area of 0.9778.

(d) When increasing the sample size, the probability decreases because the variability in the sample mean decreases as we increase the sample size which we can clearly see in part (b) and (c) of the question.

(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.

8 0
2 years ago
Select the expressions that are equivalent to 8(6m).
Bas_tet [7]

\textsf{Hey there!}

\text{Select the expressions that are equicalent to 8(6m)}\downarrow

\text{We DON'T have any like terms so we CAN'T pair them off!}

\mathsf{8(6m) \rightarrow 8 \times 6 \times \ m}\\\mathsf{8\times 6 = 48}}\\\mathsf{48\times\  m = 48m}\\\\\\\boxed{\boxed{\mathsf{Answer: \bf{48m}}}}\checkmark

\textsf{Good luck on your assignment and enjoy your day!}

~\dfrac{\frak{LoveYourselfFirst}}{:)}

4 0
2 years ago
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