Answered: 1.Find the probabilities for the…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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1.Find the probabilities for the following two situations (use 3 decimal places):

(a) Assume that IQ scores are normally distributed, with a standard deviation of 12 points and a mean of 100 points. If 110 people are chosen at random, what is the probability that the sample mean of IQ scores will not differ from the population mean by more than 2 points?

(b) A large tank of fish from a hatchery is being delivered to a lake. The hatchery claims that the mean length of fish in the tank is 15 inches, and the standard deviation is 8 inches. A random sample of 36 fish is taken from the tank. Let x be the mean sample length of these fish. What is the probability that x is within 0.5 inch of the claimed population mean?

2.What is the z-score of x = 3,

if it is 0.188 standard deviations to the left of the mean? (Enter an exact number as an integer, fraction, or decimal.)
z = .......

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Expert Solution

Part 1 (a)

Note: Since you have posted multiple independent questions in the same request, we will solve the first question for you. To get the remaining questions solved, please repost other questions independently.

$Given mean IQ (μ) = 100 I . Q Standard deviation (σ) = 12 Probability (IQ score will not differ from μ, by more than 2 points) = P (100 - 2 \leq X \leq 100 + 2) = P (98 \leq X \leq 102) For X = 98 Z = \frac{X - μ}{σ} = \frac{98 - 100}{12} = - 0.1666 And for X = 102 Z = \frac{X - μ}{σ} = \frac{102 - 100}{12} = 0.1666 Now, P (98 \leq X \leq 102) = P (- 0.166 \leq Z \leq 0.166) = P (- 0.166 \leq Z \leq 0) + P (0 \leq Z \leq 0.166) (Symmetric of Normal curve) = 0.0636 + 0.0636 (Area table of Normal Curve) = 0.1272$

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