What is the quotient of 22,389 and 13?

Suppose a geyser has a mean time between eruptions of 72 minutes. Let the interval of time between the eruptions be normally dis

nikitadnepr [17]

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 = the interval of time between the eruptions

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 = 0.3336

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 $\bar X$ = sample mean time between the eruptions

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 = 0.0582

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 $\bar X$ = sample mean time between the eruptions

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 = 0.0055

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.

6 0

3 years ago

Help plz I need to fin before today help

Alik [6]

Answer:

BD = 12.1 (nearest tenth)

Step-by-step explanation:

∆ABC is an isosceles triangle, since it has two equal sides, AB and BC. Also, this means that <BAD and <BCD = 60° each.

BD divides ∆ABC into two equal parts.

Apply trigonometric ratio to find BD.

Reference angle = <BAD = 60°

Adjacent = AD = 7

Opposite = BD

Thus, we would have:

tan 60 = opp/adj

Tan 60 = BD/7

7*Tan 60 = BD

12.1 = BD

BD = 12.1 (nearest tenth)

8 0

3 years ago

John and Andrew have £3.40 between them. John has £1.20 more than Andrew. John has £u and Andrew £v

Dahasolnce [82]

In the given question, there are several information's of immense importance and they can be used to find the necessary answers. It is already given that John and Andrew have 3.40 pound together. It is also given that John has 1.20 pound more than Andrew. It is also assumed that John has"u" pound and Andrew has "v" pounds.
Then we can write the two equations as
u + v = 3.40
u = v + 1.20
To find the values of u and v, we can replace the u in the first equation with the value of u in the second equation. Then
u + v = 3.40
(v + 1.20) + v = 3.40
2v + 1.20 = 3.40
2v = 3.40 - 1.20
2v = 2.2
v = 2.2/2
= 1.1
Now we replace the value of v in the first equation to find the value of u.
u + v = 3.40
u + 1.1 = 3.40
u = 3.40 - 1.1
u = 2.3

6 0

3 years ago

Plot the point (0,6), (3,15.8) and (9.5,0) using the graphing tool, and find the function

gladu [14]

I know the Answer I dont know how to show you. do you?

3 0

3 years ago

Read 2 more answers

How do I find the indicated derivative?

sergey [27]

Multiply the fraction by both things in the parenthesis.

8 0

3 years ago