Math Problem Statement

In each case, determine the value of the constant c that makes the probability statement correct. (Round your answers to two decimal places.)

(a) Φ(c) = 0.9850

(b) P(0 ≤ Z ≤ c) = 0.2939

(c) P(c ≤ Z) = 0.1292

(d) P(−c ≤ Z ≤ c) = 0.6528

(e) P(c ≤ |Z|) = 0.0128

Solution

We need to determine the value of the constant cc that makes each probability statement correct. For these types of problems, we are dealing with the standard normal distribution ZZ, which has a mean of 0 and a standard deviation of 1. We can use the cumulative distribution function (CDF) Φ(c)\Phi(c), the properties of the standard normal distribution, and tables or technology to find the values of cc.

(a) Φ(c)=0.9850\Phi(c) = 0.9850

Φ(c)\Phi(c) represents the cumulative distribution function for a standard normal distribution, meaning Φ(c)\Phi(c) gives the probability that ZZ is less than or equal to cc.

We need to find cc such that: Φ(c)=0.9850\Phi(c) = 0.9850 Using a standard normal table or calculator, we find that: c2.17c \approx 2.17

(b) P(0Zc)=0.2939P(0 \leq Z \leq c) = 0.2939

This represents the probability that ZZ lies between 0 and cc, so we need to find cc such that: P(0Zc)=Φ(c)Φ(0)=0.2939P(0 \leq Z \leq c) = \Phi(c) - \Phi(0) = 0.2939 Since Φ(0)=0.5\Phi(0) = 0.5, we have: Φ(c)0.5=0.2939Φ(c)=0.7939\Phi(c) - 0.5 = 0.2939 \quad \Rightarrow \quad \Phi(c) = 0.7939 Using the standard normal table: c0.82c \approx 0.82

(c) P(cZ)=0.1292P(c \leq Z) = 0.1292

This represents the probability that ZZ is greater than or equal to cc. This is the complementary probability of the cumulative distribution: P(Zc)=1Φ(c)=0.1292P(Z \geq c) = 1 - \Phi(c) = 0.1292 So, we have: Φ(c)=10.1292=0.8708\Phi(c) = 1 - 0.1292 = 0.8708 Using the standard normal table: c1.13c \approx 1.13

(d) P(cZc)=0.6528P(-c \leq Z \leq c) = 0.6528

This represents the probability that ZZ lies between c-c and cc, which is: P(cZc)=2Φ(c)1=0.6528P(-c \leq Z \leq c) = 2\Phi(c) - 1 = 0.6528 Solving for Φ(c)\Phi(c): 2Φ(c)1=0.65282Φ(c)=1.6528Φ(c)=0.82642\Phi(c) - 1 = 0.6528 \quad \Rightarrow \quad 2\Phi(c) = 1.6528 \quad \Rightarrow \quad \Phi(c) = 0.8264 Using the standard normal table: c0.94c \approx 0.94

(e) P(cZ)=0.0128P(c \leq |Z|) = 0.0128

This represents the probability that ZZ is greater than or equal to cc in absolute value, meaning: P(Zc)=2P(Zc)=0.0128P(|Z| \geq c) = 2P(Z \geq c) = 0.0128 Thus: 2[1Φ(c)]=0.01281Φ(c)=0.0064Φ(c)=0.99362[1 - \Phi(c)] = 0.0128 \quad \Rightarrow \quad 1 - \Phi(c) = 0.0064 \quad \Rightarrow \quad \Phi(c) = 0.9936 Using the standard normal table: c2.48c \approx 2.48

Final answers:

  • (a) c2.17c \approx 2.17
  • (b) c0.82c \approx 0.82
  • (c) c1.13c \approx 1.13
  • (d) c0.94c \approx 0.94
  • (e) c2.48c \approx 2.48

Would you like more details on any of these solutions? Here are 5 related questions to expand the topic:

  1. How can you calculate probabilities using the standard normal distribution without a table?
  2. What is the relationship between the CDF and the probability for a normal distribution?
  3. How can we interpret the value of Φ(c)\Phi(c) in real-world applications?
  4. How does the symmetry of the normal distribution affect these types of problems?
  5. What is the meaning of P(cZ)P(c \leq |Z|) and when is it useful?

Tip: Always remember that the cumulative distribution function Φ(c)\Phi(c) gives the area under the curve from -\infty to cc, representing the probability that a standard normal variable ZZ is less than or equal to cc.

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Math Problem Analysis

Mathematical Concepts

Probability
Standard Normal Distribution
Cumulative Distribution Function (CDF)

Formulas

Φ(c): cumulative distribution function of the standard normal distribution
P(a ≤ Z ≤ b): probability that Z lies between a and b
P(Z ≥ c) = 1 - Φ(c)
P(-c ≤ Z ≤ c) = 2Φ(c) - 1

Theorems

Properties of the Standard Normal Distribution
Cumulative Distribution Function (CDF)

Suitable Grade Level

College-level Probability or Advanced High School Statistics