The time I spend waiting for the bus on any given day has a distribution with mean 4 min- utes and variance off 0.5 minutes. What is the probability that I spend more than 2 hours and 10 minutes waiting for the bus in one month (30 days)? You may assume that waiting times on different days are independent of each other. HINT: Is there a sum of random variables somewhere in here?

0.0049 = 0.49% probability that I spend more than 2 hours and 10 minutes waiting for the bus in one month.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean and standard deviation , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

n instances of a normal variable:

For n instances of a normal variable, the mean is:

Mean of 4 minutes, standard deviation of 0.5 minutes:

This means that

30 days:

What is the probability that I spend more than 2 hours and 10 minutes waiting for the bus in one month (30 days)?

2 hours and 10 minutes is 2*60 + 10 = 130 minutes, so this probability is 1 subtracted by the p-value of Z when X = 130. So

In this context, due to the 30 instances of the normal variable:

has a p-value of 0.9951.

1 - 0.9951 = 0.0049

0.0049 = 0.49% probability that I spend more than 2 hours and 10 minutes waiting for the bus in one month.

step-by-step explanation:

equation

you are asked to write the equation

  expression = 0

using the given expression. when you do that, you have

  2x^2-2x-12 = 0

factoring

you can see that each of the coefficients is even, so a factor of 2 can be removed immediately.

  2(x^2 -x -6) = 0

if you multiply the binomials (x+a)(x+b), you get

  (x +a)(x +b) = x^2 +(a+b)x +ab

this tells you the coefficient of x is the sum of two factors of the constant term. here, that means -1 is the sum of two factors of -6. we want to know those factors so we can factor the trinomial.

  -6 = -1×6 = -2×3 = -3×2 = -6×1

the factor pair that sums to -1 is {-3, 2}, so our factorization is

  2(x -3)(x +2) = 0

solutions

the solutions to this equation are the values of x that make the product zero. a product is zero only when only of the factors of the product is zero. here, that means

  x - 3 = 0

  x = 3 . . . add 3 to both sides of the equation

or

  x +2 = 0

  x = -2 . . . add -2 to both sides of the equation

the solutions to the equation are x=3 or x=-2. both of these values of x satisfy the equation—they make it be true.