Question

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? (round your z-values to 2 decimal places and final answers to 4 decimal places.)

Answer 1

The probability of selecting a 55 one-bedroom apartment is 1 and the value of z is 38.53.

Given,

Mean of rent of one bedroom apartment = $2075

The standard deviation of rent of the apartment = $300

n = 55

x' = $1985

We have to find x' to be at least $1985 which is P(x' > $1985).

The z-value is calculated from,

z = (x' - u)/√std. deviation/n

Putting the values, we get;

z = (1985 - 2075)/(√300/55)

or, z = -38.536

So, P(x' > $1985)

or, P(z > -38.5364)

This changes to:

P(z < 38.5364)

Hence the probability from the z distribution table is 1 and the value of z is 38.5364.

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