Suppose T and Z are random variables.a. If P(T > 2.98) = 0.03 and P(T < -2.98) = 0.03, obtain P(-2.98 ≤ T ≤ 2.98).b. If P(-0.53 ≤ Z ≤ 0.53) = 0.40 and also P(Z > 0.53) = P(Z < -0.53), find P(Z > 0.53).**Solution for Part a:**- To find P(-2.98 ≤ T ≤ 2.98), recognize that the total probability is 1.00.- Since P(T > 2.98) and P(T < -2.98) are both 0.03, they sum to 0.06.- Thus, P(-2.98 ≤ T ≤ 2.98) = 1 - 0.06 = 0.94.**Graphical Explanation:**- Imagine a standard normal distribution curve.- The tails beyond ±2.98 each contain 0.03 probability.- The central area, between -2.98 and 2.98, contains 0.94 probability, representing most of the data under the curve.**Solution for Part b:**- P(-0.53 ≤ Z ≤ 0.53) = 0.40 implies that 40% of the data lies within this range.- Since P(Z > 0.53) = P(Z < -0.53), these areas are equal.- The three areas must sum to 1.00, so 1.00 - 0.40 = 0.60 is divided equally into the two tails.- Therefore, P(Z > 0.53) = 0.60 / 2 = 0.30.**Graphical Explanation:**- Consider a bell-shaped curve representing the standard normal distribution.- The middle section between Z = -0.53 and Z = 0.53 covers 0.40 of probability.- The two tails outside this range, which are equal, together cover the remaining 0.60, so each tail covers 0.30.Ensure to input your answers in the designated answer box and click "Check Answer" to confirm.

a. It is given that, P(T > 2.98)= 0.03 That is, P(T ≤ 2.98)= 1- 0.03 P(T ≤ 2.98)= 0.97 P( -…

Suppose T and Z are random variables. a. If P(T> 2.98)= 0.03 and P(T< -2.98)%= 0.03, obtain P(-2.98STS2.98). b. If P(-0.53 sZs0.53)=0.40 and also P(Z>0.53) = P(Z< -0.53). Find P(Z>0.53). a. P(-2.98STS2.98) =

Suppose T and Z are random variables. a. If P(T> 2.98)= 0.03 and P(T< -2.98)%= 0.03, obtain P(-2.98STS2.98). b. If P(-0.53 sZs0.53)=0.40 and also P(Z>0.53) = P(Z< -0.53). Find P(Z>0.53). a. P(-2.98STS2.98) =

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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Concept explainers

Continuous Probability Distributions

Probability distributions are of two types, which are continuous probability distributions and discrete probability distributions. A continuous probability distribution contains an infinite number of values. For example, if time is infinite: you could count from 0 to a trillion seconds, billion seconds, so on indefinitely. A discrete probability distribution consists of only a countable set of possible values.

Normal Distribution

Suppose we had to design a bathroom weighing scale, how would we decide what should be the range of the weighing machine? Would we take the highest recorded human weight in history and use that as the upper limit for our weighing scale? This may not be a great idea as the sensitivity of the scale would get reduced if the range is too large. At the same time, if we keep the upper limit too low, it may not be usable for a large percentage of the population!

Question

Suppose T and Z are random variables.

a. If P(T > 2.98) = 0.03 and P(T < -2.98) = 0.03, obtain P(-2.98 ≤ T ≤ 2.98).

b. If P(-0.53 ≤ Z ≤ 0.53) = 0.40 and also P(Z > 0.53) = P(Z < -0.53), find P(Z > 0.53).

**Solution for Part a:**

- To find P(-2.98 ≤ T ≤ 2.98), recognize that the total probability is 1.00.
- Since P(T > 2.98) and P(T < -2.98) are both 0.03, they sum to 0.06.
- Thus, P(-2.98 ≤ T ≤ 2.98) = 1 - 0.06 = 0.94.

**Graphical Explanation:**

- Imagine a standard normal distribution curve.
- The tails beyond ±2.98 each contain 0.03 probability.
- The central area, between -2.98 and 2.98, contains 0.94 probability, representing most of the data under the curve.

**Solution for Part b:**

- P(-0.53 ≤ Z ≤ 0.53) = 0.40 implies that 40% of the data lies within this range.
- Since P(Z > 0.53) = P(Z < -0.53), these areas are equal.
- The three areas must sum to 1.00, so 1.00 - 0.40 = 0.60 is divided equally into the two tails.
- Therefore, P(Z > 0.53) = 0.60 / 2 = 0.30.

**Graphical Explanation:**

- Consider a bell-shaped curve representing the standard normal distribution.
- The middle section between Z = -0.53 and Z = 0.53 covers 0.40 of probability.
- The two tails outside this range, which are equal, together cover the remaining 0.60, so each tail covers 0.30.

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