The mean yearly rainfall in Sydney, Australia, is about 134 mm and the standard deviation is about 66 mm ("Annual maximums of," 2013). Assume rainfall is normally distributed. How many yearly mm of rainfall would there be in the top 15%? Round answer to 2 decimal places.

At least 202.44 mm in the top 15%.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

How many yearly mm of rainfall would there be in the top 15%?

At least X mm.

X is the 100-15 = 85th percentile, which is X when Z has a pvalue of 0.85. So X when Z = 1.037.

At least 202.44 mm in the top 15%.

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b

4.08%

step-by-step explanation: