Business Statistics Questions on probability?

The scores on a pinball machine located in the Student Center are normally distributed with a mean of 200,000 points and a standard deviation of 40,000 points.

1.What is the probability that someone will score between 200,000 points and 220,000 points?



2.What is the probability that someone will score between 180,000 points and 220,000 points?



3.What is the probability that someone will score between 220,000 points and 260,000 points?



4.What is the probability that someone will score more than 260,000 points?



5.What is the probability that someone will score less than 220,000 points?

for any normal random variable you can translate into standard units and use the standard normal cdf tables to get the solutions.

Let X ~ Normal(渭 = 200000, 蟽 = 40000)

(1)

P(200000 < X < 220000)
= P( (200000 - 200000) / 40000 < Z < (220000 - 200000) / 40000)
= P(0 < Z < 0.5)
= P(Z < 0.5) - P(Z < 0)
= 0.6914625 - 0.5000
= 0.1914625


(2)

P(180000 < X < 220000)
= P(-0.5 < Z < 0.5)
= P(Z< 0.5) - P(Z < -0.5)
= 0.3829249

(3)

P(220000 < X < 260000)
= P(0.5 < Z < 1.5)
= P(Z < 1.5) - P(Z < 0.5)
= 0.2417303


(4)

P(X > 260000)
= P(Z > (260000 - 200000) / 40000)
= P(Z > 1.5)
= 1 - P(Z < 1.5)
= 0.0668072

(5)

P(X < 220000)
= P(Z < 0.5)
= 0.6914625

P(X < 200,000.0) = P((x - 200,000) / 40,000.0) < (200,000.0 - 200,000) / 40,000.0) = P(Z < 0.00) = 0.5000
P(X < 220,000.0) = P((x - 200,000) / 40,000.0) < (220,000.0 - 200,000) / 40,000.0) = P(Z < 0.50) = 0.6915

P(200,000.0 < x < 220,000.0) = 0.6915 - 0.5000 = 0.1915

P(X < 180,000.0) = P((x - 200,000) / 40,000.0) < (180,000.0 - 200,000) / 40,000.0) = P(Z < -0.50) = 0.3085
P(X < 220,000.0) = P((x - 200,000) / 40,000.0) < (220,000.0 - 200,000) / 40,000.0) = P(Z < 0.50) = 0.6915

P(180,000.0 < x < 220,000.0) = 0.6915 - 0.3085 = 0.3830


P(X < 220,000.0) = P((x - 200,000) / 40,000.0) < (220,000.0 - 200,000) / 40,000.0) = P(Z < 0.50) = 0.6915
P(X < 260,000.0) = P((x - 200,000) / 40,000.0) < (260,000.0 - 200,000) / 40,000.0) = P(Z < 1.50) = 0.9332

P(220,000.0 < x < 260,000.0) = 0.9332 - 0.6915 = 0.2417

P(X > 260,000.0) = 1 - 0.9332 = 0.0668

P(X < 220,000.0) = P((x - 200,000) / 40,000.0) < (220,000.0 - 200,000) / 40,000.0) = P(Z < 0.50) = 0.6915

I don't know if you want more than raw calculations, but presuming you do...

When you are given a normal distribution, then you need to do what's called "normalizing" the distribution to the "standard" normal distribution. A "standard" normal distribution has a mean = 0 and standard deviation = 1. (In the old days, it was necessary to look up normal distribution function values in tables. The tables showed only the "standard" values, so it was necessary to "normalize" the distribution in order to use the tables. These days modern calculators can compute non-standard normal distributions, but you learn nothing by just punching a bunch of numbers into a calculator.)

Let's call your hypothetical X. Given a normally distributed function X with mean = 碌 and standard deviation = 蟽 (usually denoted N(碌, 蟽), then if you define another distribution (let's call it Z) as
Z = ( (X - 碌) / 蟽 )
then Z is N(0,1).

So, to apply it to your first problem:

P(200,000 < X < 220,000) =
P( (200,000-200,000)/40,000 < Z < (220,000-200,000)/40,000)

= P( 0 < Z < 0.5)
= P(Z < 0.5) - P(Z < 0)

We can now look these up in a table of N(0,1) values.
P(Z < 0.5) = 0.6915. P( Z < 0) = 0.5
P(0 < Z < 0.5) = 0.1915

So, P(200,000 < X < 220,000) = 0.1915.