Question

The mean of a normal distribution is 400 pounds. The standard deviation is 10 pounds. What is the probability of a weight between 415 pounds and the mean of 400 pounds

Answer 1

The probability is
`P(x_1 \le X \le x_2 ) = 0.4332`

Explanation:

From the question we are told that

The population mean is
`\mu = 400`

The standard deviation is
`\sigma = 10`

The considered values are
`x_1 = 400 \to x_2 = 415`

Given that the weight follows a normal distribution

i.e
`\approx X (\mu , \sigma )`

Now the probability of a weight between 415 pounds and the mean of 400 pounds is mathematically as

`P(x_1 \le X \le x_2 ) = P((x_1 - \mu )/(\sigma ) \le (X - \mu )/(\sigma ) \le (x_2 - \mu )/(\sigma ) )`

So
`(X - \mu )/(\sigma )` is equal to Z (the standardized value of X )

Hence we have

`P(x_1 \le X \le x_2 ) = P((x_1 - \mu )/(\sigma ) \le Z \le (x_2 - \mu )/(\sigma ) )`

substituting values

`P(x_1 \le X \le x_2 ) = P((400 - 400 )/(10 ) \le Z \le (415 - 400)/(415 ) )`

`P(x_1 \le X \le x_2 ) = P(0\le Z \le 1.5 )`

`P(x_1 \le X \le x_2 ) = P( Z < 1.5) - P( Z < 0)`

From the standardized normal distribution table
`P( Z< 1.5) = 0.9332` and

`P( Z < 0) = 0.5`

So

`P(x_1 \le X \le x_2 ) = 0.9332 - 0.5`

`P(x_1 \le X \le x_2 ) = 0.4332`

NOTE : This above values obtained from the standardized normal distribution table can also be obtained using the P(Z) calculator at (calculator dot net).