Answer to Question #118630 in Statistics and Probability for Michael

Question

For a mean of 153 and a standard deviation of 25, find P(130 ≤ x ≤ 170 )

Answer 1

When the distribution is not mentioned, it is conventionally assumed to be Normal.

Here the distribution of X has not been mentioned, so we assume

X ~ N(μ = 153, σ² = 25²)

Then by the property of Normal distribution,

Z = $\frac{x-\mu}{\sigma}$​ ~ N(0, 1) i.e. Z is a standard normal variable

Therefore, P(130 ≤ x ≤ 170)

= P($\frac{130-153}{25}\leq\frac{x-153}{25}\leq\frac{170-153}{25}$)

= P(- 0.92 ≤ Z ≤ 0.68)

= P(Z ≤ 0.68) - P(Z < - 0.92)

= $\Phi$(0.68) - $\Phi$(- 0.92)

= 0.7517 - 0.1788 = 0.5729

Answer: The value of P(130 ≤ x ≤ 170) is 0.5729.