In a certain section of Southern California, the distribution of monthly rent for a one-bedroom apartment has a mean of $2,075 and a standard deviation of $300. The distribution of the monthly rent does not follow the normal distribution. In fact, it is positively skewed. What is the probability of selecting a sample of 55 one-bedroom apartments and finding the mean to be at least $1,985 per month

Probability is 1

Step-by-step explanation:

We are given;

mean;μ = $2,075

Standard deviation;σ = $300

n = 55

x' = $1,985

Now, we want to find x' to be at least $1,985 which is P(x' > $1,985).

The z-value is calculated from;

z = (x' - μ)/(√σ/n)

Plugging in the relevant values;

z = (1985 - 2075)/(√300/55)

z = -38.536

So, P(x' > $1,985) = P(z > -38.536)

This transforms to;

P(z < 38.536)

Probability from z distribution table is 1

the cat is blue because the probability of extension towards the prefix of the diameter of the circle could only mathematically equal to the sum of the roots of the carrots from   your garden.

step-by-step explanation: