Let X∼N(6,4).Find the probabilities P(5&lt;X&lt;7).

Step 1 Introduction: The normal probability is a type of continuous probability distribution that can take random values. The normal distribution is determined by the two parameters - the population mean (μ) and population variance (σ2). It is symmetric with respect to its mean. Given information: X∼N(6,4) Therefore, μ=6 σ2=4 Step 2 P(5&lt;X&lt;7) is computed as follows: P(5&lt;X&lt;7)=P(X&lt;7)−P(x&lt;5) =P(X−μσ2&lt;7−μσ2)−P(X−μσ2&lt;5−μσ2) =P(Z&lt;7−64)−P(Z&lt;5−64) =P(Z&lt;0.5)−P(Z&lt;−0.5) =P(Z&lt;0.5)−[1−P(Z&lt;0.5)] =0.69146−(1−0.69146) =0.69146−0.30854=0.38292 Therefore, P(5&lt;X&lt;7)=0.3829

Key to "Let X∼N(6,4).Find the probabilities P(5<X<7)."

Annette Arroyo

Answered question

2020-11-16

Let $X \sim N (6, 4)$ .Find the probabilities $P (5 < X < 7)$ .

Answer & Explanation

casincal

Skilled2020-11-17Added 82 answers

Step 1
Introduction:
The normal probability is a type of continuous probability distribution that can take random values. The normal distribution is determined by the two parameters - the population mean $(μ)$ and population variance $(σ^{2})$ . It is symmetric with respect to its mean.
Given information:
$X \sim N (6, 4)$
Therefore,
$μ = 6$
$σ^{2} = 4$
Step 2
$P (5 < X < 7)$ is computed as follows:
$P (5 < X < 7) = P (X < 7) - P (x < 5)$
$= P (\frac{X - μ}{\sqrt{σ^{2}}} < \frac{7 - μ}{\sqrt{σ^{2}}}) - P (\frac{X - μ}{\sqrt{σ^{2}}} < \frac{5 - μ}{\sqrt{σ^{2}}})$
$= P (Z < \frac{7 - 6}{\sqrt{4}}) - P (Z < \frac{5 - 6}{\sqrt{4}})$
$= P (Z < 0.5) - P (Z < - 0.5)$
$= P (Z < 0.5) - [1 - P (Z < 0.5)]$
$= 0.69146 - (1 - 0.69146)$
$= 0.69146 - 0.30854 = 0.38292$
Therefore,
$P (5 < X < 7) = 0.3829$

Do you have a similar question?

Recalculate according to your conditions!

New Questions in College Statistics

Didn't find what you were looking for?