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STatiana [176]
3 years ago
8

The probability that an event will occur is 95%. Which of these best describes the likelihood of the event occurring?

Mathematics
2 answers:
pentagon [3]3 years ago
7 0

Answer: A. Likely

I took a test and got it right.

AfilCa [17]3 years ago
4 0

Answer:

Likely

First option is correct.

Step-by-step explanation:

The probability for occurring the event = 95% = 0.95

Let us understand different types of events:-

  • Certain events: If the probability of occurring an even is 1, then it is a certain event.
  • Impossible events: If the probability of occurring an even is 0, then it is a impossible event.
  • Likely events: If the probability of occurring an even is close to 1, then it is a likely event.
  • Unlikely events: If the probability of occurring an even is close to 0, then it is a unlikely event.

Now, the probability for occurring the event = 95% = 0.95, which is close to 1.

Hence, we can conclude that the event is likely.

First option is correct.

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Answer:

(a) The probability that a randomly selected time interval between eruptions is longer than 82 ​minutes is 0.3336.

(b) The probability that a random sample of 13-time intervals between eruptions has a mean longer than 82 ​minutes is 0.0582.

(c) The probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 ​minutes is 0.0055.

(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.

(e) The population mean must be more than 72​, since the probability is so low.

Step-by-step explanation:

We are given that a geyser has a mean time between eruptions of 72 minutes.

Also, the interval of time between the eruptions be normally distributed with a standard deviation of 23 minutes.

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

So, X ~ N(\mu=72, \sigma^{2} =23^{2})

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 = 72 minutes

           \sigma = standard deviation = 23 minutes

Now, the probability that a randomly selected time interval between eruptions is longer than 82 ​minutes is given by = P(X > 82 min)

       P(X > 82 min) = P( \frac{X-\mu}{\sigma} > \frac{82-72}{23} ) = P(Z > 0.43) = 1 - P(Z \leq 0.43)

                                                           = 1 - 0.6664 = <u>0.3336</u>

The above probability is calculated by looking at the value of x = 0.43 in the z table which has an area of 0.6664.

(b) Let \bar X = <u><em>sample mean time between the eruptions</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 = 72 minutes

           \sigma = standard deviation = 23 minutes

           n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between eruptions has a mean longer than 82 ​minutes is given by = P(\bar X > 82 min)

       P(\bar X > 82 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{82-72}{\frac{23}{\sqrt{13} } } ) = P(Z > 1.57) = 1 - P(Z \leq 1.57)

                                                           = 1 - 0.9418 = <u>0.0582</u>

The above probability is calculated by looking at the value of x = 1.57 in the z table which has an area of 0.9418.

(c) Let \bar X = <u><em>sample mean time between the eruptions</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 = 72 minutes

           \sigma = standard deviation = 23 minutes

           n = sample of time intervals = 34

Now, the probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 ​minutes is given by = P(\bar X > 82 min)

       P(\bar X > 82 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{82-72}{\frac{23}{\sqrt{34} } } ) = P(Z > 2.54) = 1 - P(Z \leq 2.54)

                                                           = 1 - 0.9945 = <u>0.0055</u>

The above probability is calculated by looking at the value of x = 2.54 in the z table which has an area of 0.9945.

(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.

(e) If a random sample of 34-time intervals between eruptions has a mean longer than 82 ​minutes, then we conclude that the population mean must be more than 72​, since the probability is so low.

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3 years ago
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