Hi hi. i have a legit question now.
if the mean is 52$ and the standard deviation is 3$ how do you find the z score? my problems for this are
a. less than 50
b. more than 59
c. more than 40
and d. between 48 and 57$
but i'm a little unsure on how to find these percentages without the x value.

A. 25.14%

B. 0.99%

C. 100%

D. 86.07%

Step-by-step explanation:

μ = 52, σ = 3

Each problem gives you an x value and asks you to find a probability.  First calculate the z score(s), then look up in a table or use a calculator.

z = (x − μ) / σ

A. Find P(x < 50)

z = (50 − 52) / 3

z = -0.67

P(z < -0.67) = 0.2514

B. Find P(x > 59)

z = (59 − 52) / 3

z = 2.33

P(z > 2.33) = 1 − 0.9901 = 0.0099

C. Find P(x > 40)

z = (40 − 52) / 3

z = -4

P(z > -4) = 1 − 0 = 1

D. Find P(48 < x < 57)

z₁ = (48 − 52) / 3

z₁ = -1.33

z₂ = (57 − 52) / 3

z₂ = 1.67

P(-1.33 < z < 1.67) = 0.9525 − 0.0918 = 0.8607

I used a z score table.  For more accurate answers, use a calculator.

answer:

1. 0.04

2. 0.15

step-by-step explanation:

i just finished that assignment for properties of probability distributions.

im not sure

step-by-step explanation: