Question

Use the normal distribution of fish lengths for which the mean is 10 inches and the standard deviation is 2 inches. Assume the variable x is normally distributed. left parenthesis a right parenthesis What percent of the fish are longer than 11 ​inches? left parenthesis b right parenthesis If 500 fish are randomly​ selected, about how many would you expect to be shorter than 8 ​inches? left parenthesis a right parenthesis Approximately nothing​% of fish are longer than 11 inches. ​(Round to two decimal places as​ needed.) left parenthesis b right parenthesis You would expect approximately nothing fish to be shorter than 8 inches. ​(Round to the nearest​ fish.)  

Answer 1

Solution:
Given in the question
the variable x is normally distributed with
Mean () = 10
Standard deviation () = 2
Solution(a)
We need to calculate percentage of fish are longer than 11 inches i.e. P(X>11) = ?
Here we will use standard normal distribution table, First we will calculate Z-score which can be calculated as
Z-score. = (X-)/ = (11-10)/2 = 0.5
From Z table we found p-value
P(X>11) = 0.3085
So there is 30.85% of fish are longer than 11 inches.
Solution(b)
We need to calculate percentage of fish are Shorter than 8 inches i.e. P(X<8) = ?
Here we will use standard normal distribution table, First we will calculate Z-score which can be calculated as
Z-score. = (X-)/ = (8-10)/2 = -1
From Z table we found p-value
P(X<8) = 0.1587
So there is 15.87% of fish are Shorter than 8 inches.
So Out of 500 fish, there is (0.1587*500) = 79 fish are shorter than 8 inches.