Question

Long term weather records in Scotland show that on average 80% of days are cloudy in the month of November.
Using the normal distribution approximation, find the probability of November having fewer than 25 cloudy days.
Take November to have 30 days.

Answer 1

@order posted this question previously and said the answer was a probability of 0.59.
I hope I have remembered everthing correctly :)

Answer 2

Here is my attempt at a solution.
The average number of cloudy days = 30 * 0.8 = 24 days.
The maximum number of cloudy days = 30
The minimum number of cloudy days = 18 based on a normal distribution which is symmetrical.
The spread of the distribution is 30 - 18 = 12 days.
Nearly all of the area of the distribution is included within a distance of +- 3 standard deviations from the mean.
\[6\sigma=12days\]
\[\sigma=2days\]
When approximating discrete distributions with the normal distribution it is necessary to make a 'continuity correction' of half a unit at the ends of the relative frequency histogram.
In this case when transforming the variable X < 25 into a standardised normal variable Z the value of 24.5 must be used for X.
\[Z=\frac{X-mu}{\sigma}=\frac{24.5-24.0}{2}=0.25\]
Using Z = 0.25 in a standard normal distribution table gives a probability of fewer than 25 cloudy days in November = 0.5974

Answer 3

why aren't you using the formula for the standard deviation of the binomial to give you \(\sigma\)?

Answer 4

@Zarkon can you please explain further. What would the result be for sigma?

Answer 5

\[\sqrt{n\cdot p\cdot q}\]
\[=\sqrt{30\times .8\times.2}\]

Answer 6

@Zarkon thank you very much.The more exact value for sigma when used in the calculation of Z leads to a probability of fewer than 25 cloudy days in November = 0.59

Answer 7

Thanks so much for helping me understand this! I really appreciate it :D

Answer 8

You're very welcome :)