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vivado [14]
3 years ago
11

2203

Mathematics
1 answer:
ale4655 [162]3 years ago
5 0

Answer:

sorry im just answering this cuz need points

Step-by-step explanation:

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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
3 years ago
How long will it take an airplane to travel 1,250 kilometers if it is traveling at 150km/hr
Mila [183]
Time = distance/speed
So all you have to do is plug the numbers in and you get the answer
Time = 1,250 kilometres/150 kilometres per hour
Time = 8.333 hours
7 0
3 years ago
Read 2 more answers
Look at the table. Make a conjecture about the sum of the first 20 positive even numbers.
Ludmilka [50]
It will be 20*21 based on the pattern in the chart
3 0
3 years ago
Read 2 more answers
Can someone help me with this ?
telo118 [61]
(w³+7)/w=(9w²+7)/9
(w³+7)9=(9w²+7)x
9w³+63=9w³+7x
63=7x
x=9
7 0
4 years ago
Read 2 more answers
The polynomial equation x^3-4x^2+2x+10=x^2-5x-3 has complex roots 3+-2i What is the other root? Use a graphing calculator and a
kirill115 [55]

Answer:

B. -1

Step-by-step explanation:

x^3-4x^2+2x+10=x^2-5x-3

We know it has 3 roots since it is a 3rd degree polynomial.

Two of the roots are (3+2i) and (3-2i)

Subtract x^2-5x-3 from both sides

x^3-4x^2+2x+10-(x^2-5x-3)=x^2-5x-3 -(x^2-5x-3)

Distribute the minus sign

x^3-4x^2+2x+10-x^2+5x+3=x^2-5x-3 -x^2+5x+3

x^3 -5x^2+7x +13 =0

Graphing this equation , we see that it crosses the x axis at x=-1

That covers the three roots, 1 real and two complex

7 0
3 years ago
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