the following table shows the ages of the patients admitted in a hospital during a year:

find the mode and the mean of the data given above. compare and interpret the two measures of central tendency.
as per the question:
modal class = 35 -45
l = 35
h = 10
f1 = 23
f0 = 21
f2 = 14

= 

= 36.8
the above given data can be represented in the form of table as below:



= 35.37
the mode of the data shows that maximum number of patients in the age group of 36.8, whereas the average age of all the patients is 35.37.
what’s your day rate. Simply using `df.mean()` will Do The Right Thing(tm) with respect to NaNs:



```
>>> df
                 A      B
DATE                     
2013-05-01  473077  71333
2013-05-02   35131  62441
2013-05-03     727  27381
2013-05-04     481   1206
2013-05-05     226   1733
2013-05-06     NaN   4064
2013-05-07     NaN  41151
2013-05-08     NaN   8144
2013-05-09     NaN     23
2013-05-10     NaN     10
>>> df.mean(axis=1)
DATE
2013-05-01    272205.0
2013-05-02     48786.0
2013-05-03     14054.0
2013-05-04       843.5
2013-05-05       979.5
2013-05-06      4064.0
2013-05-07     41151.0
2013-05-08      8144.0
2013-05-09        23.0
2013-05-10        10.0
dtype: float64

```

You can use `df[["A", "B"]].mean(axis=1)` if there are other columns to ignore. histogram  divide data o equal width o no overlapping  bar chart o skew­ pulls direction of mean o symmetric numeric summary center, spread, and location  greek letters for population  roman letters for statistics 1. mean  µ= population mean  x bar= sanple mean o ∑ xi  all over n= sample mean  k=0 2. mean vs. median a. mean used the most b. n is small when outliers occur c. use median, mean is too affected spread 1. range (max­min) a. sensitive to outliers 2. iqr a. not sensitive to outliers variability­ xi­x bar solution: in a given data, the highest frequency is 41, which lies in the interval 1000-15000.   
 here, l=10000, `f\_(m)=41=26, f\_(2)=16 and h=5000`   
 `"mode""=l+(f\_(m)-f\_(1))/(2f\_(m)-f\_(1)-f\_(2))xxh`   
 `=1000+((41-26)/(2xx41-26-16))xx5000`   
 `=1000+((15)/(82-42))xx5000`   
 `=1000+((15)/(40))xx5000`   
 `=1000+15xx125=10000+1875=11875`   
 hence, the modal income is 11875. I answered your question, but I'll go further here for you. My FASTING glucose was 174, and they told me it was possibly diabetes. They recommended followup with my doctor within 2 weeks.   

  

The doctor will perform an A1C test - that will indicate whether you do or don't have diabetes. It's a simple finger stick that takes 5 minutes to read.   

  

Satisfied now? all laboratory results need to be interpreted in the clinical context and the doctor who ordered the tests is usually in the best position to do that. having said that, any muscular activity will raise ck level. you may visit this site for information on this topic: https://www.mayomedicallaboratories.com/test-catalog/clinical+and+interpretive/8336 see more... you seem to be missing some information about e.if e is the midpoint between d and f, thene is (1,1) take half of (4-2) = 1 and half of (5-3) = 1e alphabetically does come between d and f.just average the x values and average the y values of the two points d and for plot them graphically, carefully, maybe on some graph paper, and you'llvisually see the midpoint is at (1,1) sylvia frank chlh 274 homework problems chapter 3 1. a. crude death rate: total populace death in a specific span-time at the mean year. b. age-specific rate: total incidence occurred in a specific-age group of the populace in a specific span time. c. cause-specific rate: total death of a populace from a specific-cause in a specific span-time at the mean year. d. proportional mortality ratio: total death within a populace due to a specific-cause divided by the total death in the populace e. maternal mortality rate: incidence of death because of birth of a child or due to pregnancy associated complications. f. infant mortality rate: rate of event of newborn death over a year by birth of newborn over a same interval. g. neonatal mortality more like retain oxygen intake /efficiency. hii..

according to me,the cut off will be around 90-95 marks for general category,50-55 marks for obc category,35-40 marks for sc category and 25-30 marks for st category.so it is difficult to be predict now.

please see the table below for previous years trends:-

 

jee main cut off trends (2013-2017)

| s.no. | category | 2017 | 2016 | 2015 | 2014 | 2013 |
| 1 | general | 81 | 100 | 105 | 115 | 113 |
| 2 | obc-ncl | 49 | 70 | 70 | 74 | 70 |
| 3 | sc | 32 | 52 | 50 | 53 | 50 |
| 4 | st | 27 | 48 | 44 | 47 | 45 |