Math Problem Statement

1. If the amount of cosmic radiation to which a person is exposed while flying by jet across the US is a random variable with mean 4.35 mrem and standard deviation 0.59 mrem, find the probability that the amount of cosmic radiation to which a person will be exposed on such a flight is (a) between 4.00 and 5.00 mrem (b) at least 5.50 mrem. 2. If a random variable has the probability density function f(x) = k * e^(-x/3) for x > 0 and 0 for x <= 0, find the value of k.

Solution

Let's solve the problems one by one.

Problem 1:

Given:

  • The mean (μ\mu) = 4.35 mrem
  • The standard deviation (σ\sigma) = 0.59 mrem
  • The amount of cosmic radiation is modeled as a normally distributed random variable.

We need to find the probabilities for the following cases:

(a) The probability that the radiation is between 4.00 and 5.00 mrem.

To find this probability, we convert the raw values to standard normal values (z-scores) using the formula: z=xμσz = \frac{x - \mu}{\sigma}

  1. For x=4.00x = 4.00: z1=4.004.350.590.59z_1 = \frac{4.00 - 4.35}{0.59} \approx -0.59

  2. For x=5.00x = 5.00: z2=5.004.350.591.10z_2 = \frac{5.00 - 4.35}{0.59} \approx 1.10

We then find the probabilities corresponding to z1z_1 and z2z_2 using standard normal distribution tables or a calculator:

  • P(z<0.59)0.2776P(z < -0.59) \approx 0.2776
  • P(z<1.10)0.8643P(z < 1.10) \approx 0.8643

The probability that xx is between 4.00 and 5.00 mrem is: P(4.00<x<5.00)=P(z<1.10)P(z<0.59)0.86430.2776=0.5867P(4.00 < x < 5.00) = P(z < 1.10) - P(z < -0.59) \approx 0.8643 - 0.2776 = 0.5867

(b) The probability that the radiation is at least 5.50 mrem.

First, find the z-score for x=5.50x = 5.50: z3=5.504.350.591.95z_3 = \frac{5.50 - 4.35}{0.59} \approx 1.95

The probability that zz is at least 1.95 is: P(z>1.95)=1P(z<1.95)10.9744=0.0256P(z > 1.95) = 1 - P(z < 1.95) \approx 1 - 0.9744 = 0.0256

Problem 2:

Given:

  • The probability density function is f(x)=kex/3f(x) = k e^{-x/3} for x>0x > 0, and f(x)=0f(x) = 0 for x0x \leq 0.

We need to find the value of kk.

To satisfy the properties of a probability density function, the integral of f(x)f(x) over the entire range must equal 1: 0kex/3dx=1\int_{0}^{\infty} k e^{-x/3} \, dx = 1

Calculating the integral: k0ex/3dx=1k \int_{0}^{\infty} e^{-x/3} \, dx = 1 The integral of ex/3e^{-x/3} is 3ex/3-3 e^{-x/3}, so: k[3ex/3]0=1k \left[ -3 e^{-x/3} \right]_{0}^{\infty} = 1 k[0(3)]=1k \left[ 0 - (-3) \right] = 1 3k=13k = 1 Thus, k=13k = \frac{1}{3}.

Would you like more details, or do you have any further questions?

Related Questions:

  1. What is a z-score and how is it used in probability calculations?
  2. How do you find probabilities using the standard normal distribution table?
  3. What is the integral of a function and why is it used in probability density functions?
  4. How can the properties of a normal distribution be applied to solve real-world problems?
  5. What is the significance of standardizing a normal distribution?

Tip:

When dealing with probabilities for normal distributions, converting raw scores to z-scores simplifies the calculations by using standard normal distribution tables.

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Math Problem Analysis

Mathematical Concepts

Normal Distribution
Z-Scores
Probability Density Functions

Formulas

z = (x - μ) / σ
Integral of exponential functions
Properties of probability density functions

Theorems

Standard Normal Distribution Theorem
Fundamental Theorem of Calculus

Suitable Grade Level

Undergraduate Level or Advanced High School