The daily dinner bills in a local restaurant are normally distributed with a mean of $28 and a standard deviation of $6. What are the minimum value of the bill that is greater than 95% of the bills?

The minimum value of the bill that is greater than 95% of the bills is $37.87.

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:

What are the minimum value of the bill that is greater than 95% of the bills?

This is the 95th percentile, which is X when Z has a pvalue of 0.95. So X when Z = 1.645.

The minimum value of the bill that is greater than 95% of the bills is $37.87.

let x and 400 - x be the two numbers.  

==> 0.80x + 0.85(400 - x) = 400 - 68 from the statement of the problem.  

< ==> 0.80x + 0.85(400 - x) = 400 - 68  

==> 340 - 0.05x = 332  

< ==> -0.05x = -8 ==> x = 160, the first number ==> 400 - x = 240, the second number.

mr. india