Question

1.Assume that a distribution has a mean of 21 and a standard deviation of 4. What percentage of the values in the distribution do we expect to fall between 17 and 21?

2. Suppose that the luxury sales tax rate in a foreign country is 27%. A very wealthy socialite bought a diamond tiara for $175,000. How much tax does she pay?

Answer 1

Q1 : 34.13%

From properties of Normal Distribution that approximately 68% of the values lie within 1 Standard Deviation on either side of mean. Given that mean is 21 and standard deviation is 17 which is 1 standard deviation below mean, approximately 68%/2 = 34% values would lie between 17 and 21

We can find the exact percentage as below

We know that 50% of the values would lie below the mean and 50% would lie above the mean. The mean in this case is 21. Hence we know that 50% of the values would lie below the mean.

% of values which lie between 17 and 21 = % of values below 21 - % of values below 17

% of values which lie between 17 and 21 = 0.5 - % of values below 17

To find % of values below 17 we need to find P(x<17)

$P(x<17) = P(z<\frac{17-21}{4}) = P(z < -1) = 0.1587$

Hence % of values which lie between 17 and 21 = 0.5 - 0.1587 = 0.3413 = 34.13%

Q2 : It is given that the socialite paid $175000 for the dress. The total amount which is paid includes the tax. Hence is the total amount before tax is X and tax rate is 27%, we have the below

X+27%*X = 175000

1.27X = 175000

X = 175000/1.27 = 137795.28

Tax is 27% of X. Hence tax = 0.27*137795.28 = 37204.72

Hence total tax paid is $37204.72