Question

A biologist wants to see if rats running in a maze use some kind of decision process on how to proceed when they come to an intersection. To do so she decides to simulate the situation by assuming the decisions are random, so she creates a "virtual rat" that at each intersection tosses a coin to decide which way to go. She lets the virtual rat run the maze 1,000 times, and finds the average length run is 14. 8 feet, with a standard deviation of 2. 4 feet. She then lets a real rat run the maze, and measures the length run to be 13 feet. Explain what conclusion the biologist can make.

Answer 1

She then lets a real rat run the maze and measures the long run to be 13 feet.

What is a normal distribution?

A function that represents the distribution of many random variables as a symmetrical bell-shaped graph.

We first need to find the z-score which is given by applying the following formula:

z-score = (X-μ) ÷ σ

Where: μ is the average, σ is the standard deviation, X is the sampled value.

Therefore,

z-score = (13-14.8) ÷ 2.4

z-score = -0.75

Finding the probability, or the p-value by looking upon the z-table

P(Z<z) = P(Z<-0.75) = 0.2266

The conclusion that the biologist could make is that out of 1000 virtual rats, there are 0.2266×1000 = 226.6 ≈ 227 virtual rats that run the maze with a length of 13 feet.

Hence, She then lets a real rat run the maze and measures the long run to be 13 feet.

To know more about Normal distribution click the link given below.

https://brainly.com/question/13759327