How much a person can carry in train?

Answers

Answer 1

Answer:

two personal items, 25 lbs. (12 kg) and 14 x 11 x 7 inches each, and two carry-on items, 50 lbs. (23 kg) and 28 x 22 x 14 inches each

Explanation: I've been on a train before

Answer 2

Passengers that are known to be travelling over all Indian Railways are said to be permitted to carry luggage free of charge with them in the compartment up to about 70 Kgs.

How much a person can carry in train?

It does depends on the company that runs the train, what type of ticket you buy, and any rules about how much baggage you can bring.

Most train companies have rules about how much luggage passengers are allowed to bring with them. These rules tell you how heavy and big your bags are allowed to be when you bring them on a plane or check them in.

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Related Questions

use the sample information X =36, 0=7,n=20 to calculate the following confidence interval for u assuming the sample is from a normal population

Answers

The 90 percent confidence interval for the population mean (μ) is (33.423, 38.577).

How to calculate the value

Confidence Interval = X ± Z * (s/√n)

Where:

s is the sample standard deviation.

Given:

X = 36

s = 7

n = 20

For a 90 percent confidence level, the Z-score is 1.645 (corresponding to the 0.05/2 critical values from the standard normal distribution table).

Plugging in the values:

Confidence Interval = 36 ± 1.645 * (7/√20)

Confidence Interval = 36 ± 1.645 * (7/√20)

= 36 ± 1.645 * (7/4.472)

= 36 ± 1.645 * 1.567

= 36 ± 2.577

Therefore, the 90 percent confidence interval for the population mean (μ) is (33.423, 38.577).

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use the sample information X =36, 0=7,n=20 to calculate the following confidence interval for u assuming the sample is from a normal population (a) 90 percent confidence

mode is 34 and median Class limits 0-9 10-19 20-29 30-39 40-49 50-59 60-69 Total frequency 4 16 f? f? f? 6 4 Total 230​

Answers

Answer:

Explanation:

To find the missing frequencies in the table, we need to calculate the remaining values.

Given:

Mode = 34

Median = ?

Class limits: 0-9, 10-19, 20-29, 30-39, 40-49, 50-59, 60-69

Total frequency: 4, 16, f?, f?, f?, 6, 4

Total = 230

First, let's find the median class. The median is the middle value when the frequencies are cumulatively added up and reach half of the total frequency.

Total frequency = 230

Median class: ???

To find the median class, we calculate the cumulative frequencies by adding up the frequencies starting from the first class interval until the cumulative frequency is greater than or equal to 115 (half of the total frequency).

Cumulative frequency:

0-9: 4

10-19: 20 (4 + 16)

20-29: ??

30-39: ??

40-49: ??

50-59: ??

60-69: 230 (4 + 16 + ?? + ?? + ?? + 6 + 4)

Since the cumulative frequency of 30-39 is greater than or equal to 115 (half of the total frequency), the median class is 30-39.

Now, let's find the median value within the median class.

Median class: 30-39

Median value: ???

To find the median value within the median class, we need to know the cumulative frequency of the class just before the median class, as well as the frequency of the median class.

Cumulative frequency of the class before the median class (20-29): ??

The cumulative frequency of the class before the median class plus the frequency of the median class should be greater than 115. Let's assume the missing cumulative frequency of the class before the median class is represented by "x."

x + ? (frequency of the median class) > 115

Since the total frequency is 230, we can deduce:

230 - 115 = x + ??

So, x = 115 - ??

Now, let's calculate the median value within the median class. Since the median class is 30-39, the median value falls in the range of 30-39.

Median value within the median class: ??

To find the median value within the median class, we can use the formula:

Median value = Lower limit of the median class + (Frequency / 2 - Cumulative frequency before median class) × Class width

In this case, the lower limit of the median class is 30, and the class width is 10.

Median value within the median class = 30 + ((? / 2) - x) × 10

Now we have all the information needed to find the missing frequencies.

Please provide the value for the cumulative frequency of the class before the median class (20-29) so that we can proceed with the calculations.

Answer:

Firstly, let's calculate the missing frequency for the class interval of 30-39.
Given that the mode is 34, it means the class interval with the highest frequency contains the value 34.

From the given table, we can see that the class interval of 30-39 has a frequency of 6.
As the value 34 falls within this interval, we can assume that the frequency of this interval is also the mode, which is 34.  

Now, let's calculate the missing frequencies for the class intervals of 40-49 and 50-59.  

Given that the median is the middle value of the data, we can find the median class interval by finding the cumulative frequency just greater than half of the total frequency.  

Total frequency = 230 Half of the total frequency = 115  Cumulative frequency:

Class interval 0-9: 4

Class interval 10-19: 16 (20)

Class interval 20-29: f? (f?+20)

Class interval 30-39: 34 (f?+20+6)

Class interval 40-49: f? (f?+20+6+34)

Class interval 50-59: f? (f?+20+6+34+f?)

The cumulative frequency just greater than 115 is 126, which falls in the class interval of 40-49.

Therefore, the median class interval is 40-49.  To find the median, we can use the formula:

Median = l + ((n/2 - cf) / f) * w where, l = lower class limit of the median class n = total frequency cf = cumulative frequency just less than the median f = frequency of the median class w = width of the median class  


Using the values from the table, we can calculate the median as:

Median = 30 + ((115-20) / 34) * 10

Median = 30 + (95/34) * 10

Median = 30 + 27.94

Median ≈ 57.94  

As the median falls in the median class interval of 40-49, we can assume that the frequencies of the class intervals of 40-49 and 50-59 are equal. Let's represent the frequency of these class intervals as x.  

We know that the total frequency is 230.

Therefore, 4 + 16 + 6 + 34 + x + x + x = 230 58 + 3x = 230 3x = 172 x ≈ 57.33  Rounding off the value of x to the nearest whole number, we get x = 57.  

Hence, the missing frequencies are:

Class interval 20-29: 57

Class interval 40-49: 57

Class interval 50-59: 57

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