just wondering how you compare probabilities when asked to compare normal distributions to the histogram?? also how do you do question one (iv) (b)
in the 2018 exam https://www.nzqa.govt.nz/nqfdocs/ncea-resource/exams/2018/91267-exm-2018.pdf
also how do you calculate the mean/median of a histogram
Calculating the mean and median is much the same as you would for a list of numbers. For a histogram you would add up all the values then divide by how many in total.
For the median, since the histogram is already in order from smallest to largest, it is simply a matter of identifying the middle. Remember you want half on each side. You may have to estimate the value rather than calculate it.
Do you have a specific example of this that you want help with?
Hi, @mego! To answer the question from the assessment, you will have to solve a simple algebraic equation. You will have to find the average daily temperature if you know that the probability for the daily temperature to be below 9 degrees is 20%.
You either use your calculator or a table to find Z (remember if you use table you will have to find Z for the probability 0.3) which is equal to 0.8416. That means that 9 is 0.8416 standard deviations away (to the left) from the mean average.
So the new expected mean average temperature is 13.04 degrees.
I have been looking at your question from the 2018 paper and wanted to add an explantion about question one (iv) (b)
"(b) Show that, for any normal distribution, the inter-quartile range (IQR) is always greater than
the standard deviation. "
This question generalises normal distribution so you can answer it using the standard normal curve, where the mean is 0 and the standard deviation is 1
The interquartile range is the middle 50%
From the tables or using your graphics calculator the middle 50% on the standard curve is between -0.674 and 0.674 so a total area of 1.348
hence 1.348 > 1.
I hope this helps
please let me know if you want more explanation