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3 years ago

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The histogram to the right represents the weights (in pounds) of members of a certain high-school debate team. What is the cla

ss width? What are the approximate lower and upper class limits of the first class? The class width is_______.

1 answer:

nikitadnepr [17]3 years ago

8 0

Answer:

Class width = 20

Approximate lower class limit of the first class = 110

Approximate Upper class limit of the first class = 119

Step-by-step Explanation:

The class width of the histogram attached below can be gotten by finding the difference between successive lower class limits.

Thus, class width = 130 - 110 = 20

The approximate lower class limit of the first class is the lowest score we have in the first class = 110

The approximate upper class limit of the first class is the closest highest score that fall within the first class and is below the lower limit of the second class. Thus approximate upper class limit of the first class = 129

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Step-by-step explanation:

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1 inch = 2.54 centimeters. How many centimeters long is an 18-inch ruler?

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2 years ago

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The number of airline passengers in 1990 was 466 million. The number of passengers traveling by airplane each year has increased

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Answer:

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Step-by-step explanation:

We have been given an exponential growth formula $P(t)=466\cdot 1.035^t$ , which represents number of passengers traveling by airplane since 1990.

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$900=466\cdot 1.035^t$

$\frac{900}{466}=\frac{466\cdot 1.035^t}{466}$ $1.9313304721030043=1.035^t$

Take natural log of both sides:

$\text{ln}(1.9313304721030043)=\text{ln}(1.035^t)$

Using property $\text{ln}(a^b)=b\cdot \text{ln}(a)$ , we will get:

$\text{ln}(1.9313304721030043)=t\cdot \text{ln}(1.035)$

$0.658209129198=t\cdot 0.034401426717$

$\frac{0.658209129198}{0.034401426717}=\frac{t\cdot 0.034401426717}{0.034401426717}$

$19.1331927775=t\\\\t=19.1331927775$

This means that in the 20th year since 1990, 900 million passengers would travel by airline.

$1990+20=2010$

Therefore, 900 million passengers would travel by airline in 2010.

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2 years ago

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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.

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2 years ago