Answer:
P(X<135)=0.0228
Step-by-step explanation:
The data Lisa collected to find the number of pages per book shelf has normal distribution with mean 191 pages and standard deviation is 28 pages
We first find the z-score using:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
We substitute [tex]x=135,\mu=191,\sigma=28[/tex]
[tex]z=\frac{135-191}{28}[/tex]
[tex]z=-2.0[/tex]
We read from the standard normal distribution table -2 under 0 to obtain:
P(X<135)=0.0228
The probability that a randomly selected book has fewer than 135 pages is 0.02275
Step-by-step explanation:
To find the probability of random variable X which has a normal distribution we use the z-score
The formula of z-score is z = (X - μ)/σ , where
∵ The mean is 191 pages
∴ μ = 191
∵ The standard deviation is 28 pages
∴ σ = 28
∵ The book has fewer than 135 pages
∴ X = 135
- Substitute all of these numbers in the formula of z-score
∴ z-score = [tex]\frac{135-191}{28}[/tex]
∴ z-score = -2
Use the normal distribution table of z to find the corresponding area of the z-score
∵ The corresponding area of z = -2 is 0.02275
- The probability is the area to the left of z-score
∴ P(X < 135) = 0.02275
The probability that a randomly selected book has fewer than 135 pages is 0.02275
