Math Problem Statement

Given that x is a normally distributed variable with a mean of 60 and a standard deviation of 10, find P(x > 60), P(x < 72), P(65 < x < 82); and the waiting time x for a lecture is approximately normally distributed with a mean of 3.7 minutes and standard deviation of 1.4 minutes, find the probability that a randomly selected course has to wait for less than 2.0 minutes, and the probability that a course has to wait between 5.0 and 7.5 minutes.

Solution

Let's go through each question step by step:

Question 3:

Given that xx is a normally distributed variable with a mean of 6060 and a standard deviation of 1010, find the following probabilities:

a) P(x>60)P(x > 60)

Since xx follows a normal distribution, the probability of xx being equal to the mean (60) is exactly 0.5. The area to the right of 60 in a standard normal distribution (which has a mean of 0) is also 0.5.

So: P(x>60)=0.5P(x > 60) = 0.5

b) P(x<72)P(x < 72)

First, we calculate the z-score for 72: z=726010=1.2z = \frac{72 - 60}{10} = 1.2

We then use the standard normal distribution table to find P(z<1.2)P(z < 1.2). The value for z=1.2z = 1.2 is approximately 0.8849.

Thus: P(x<72)=0.8849P(x < 72) = 0.8849

c) P(65<x<82)P(65 < x < 82)

We need to find two z-scores:

  1. For x=65x = 65: z1=656010=0.5z_1 = \frac{65 - 60}{10} = 0.5 From the standard normal table, P(z<0.5)0.6915P(z < 0.5) \approx 0.6915.

  2. For x=82x = 82: z2=826010=2.2z_2 = \frac{82 - 60}{10} = 2.2 From the table, P(z<2.2)0.9861P(z < 2.2) \approx 0.9861.

Now, to find P(65<x<82)P(65 < x < 82): P(65<x<82)=P(z<2.2)P(z<0.5)=0.98610.6915=0.2946P(65 < x < 82) = P(z < 2.2) - P(z < 0.5) = 0.9861 - 0.6915 = 0.2946

Question 4:

The waiting time xx for a lecture to begin follows a normal distribution with a mean of 3.73.7 minutes and a standard deviation of 1.41.4 minutes.

a) Probability that a randomly selected course has to wait for less than 2.0 minutes.

Calculate the z-score for 2.0: z=2.03.71.4=1.21z = \frac{2.0 - 3.7}{1.4} = -1.21

Using the standard normal table, P(z<1.21)0.1131P(z < -1.21) \approx 0.1131.

So: P(x<2.0)=0.1131P(x < 2.0) = 0.1131

b) Probability that a randomly selected course has to wait between 5.0 and 7.5 minutes.

  1. For x=5.0x = 5.0: z1=5.03.71.4=0.93z_1 = \frac{5.0 - 3.7}{1.4} = 0.93 From the table, P(z<0.93)0.8238P(z < 0.93) \approx 0.8238.

  2. For x=7.5x = 7.5: z2=7.53.71.4=2.71z_2 = \frac{7.5 - 3.7}{1.4} = 2.71 From the table, P(z<2.71)0.9966P(z < 2.71) \approx 0.9966.

Now, to find P(5.0<x<7.5)P(5.0 < x < 7.5): P(5.0<x<7.5)=P(z<2.71)P(z<0.93)=0.99660.8238=0.1728P(5.0 < x < 7.5) = P(z < 2.71) - P(z < 0.93) = 0.9966 - 0.8238 = 0.1728

Summary:

  • 3a) P(x>60)=0.5P(x > 60) = 0.5
  • 3b) P(x<72)=0.8849P(x < 72) = 0.8849
  • 3c) P(65<x<82)=0.2946P(65 < x < 82) = 0.2946
  • 4a) P(x<2.0)=0.1131P(x < 2.0) = 0.1131
  • 4b) P(5.0<x<7.5)=0.1728P(5.0 < x < 7.5) = 0.1728

Would you like further explanations or details on any of these parts?

Here are 5 related questions for further exploration:

  1. How do you calculate the z-score for a given value in a normal distribution?
  2. What is the significance of the standard normal distribution table in solving these problems?
  3. How do you interpret probabilities in a normal distribution?
  4. How would the probabilities change if the standard deviation was smaller or larger?
  5. What happens to the probability P(x>a)P(x > a) as aa increases beyond the mean?

Tip: Always visualize the normal distribution curve when solving probability questions; it helps to understand the area under the curve corresponding to the probabilities you seek.

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Math Problem Analysis

Mathematical Concepts

Normal Distribution
Z-Score Calculation
Probability

Formulas

Z = (X - μ) / σ
P(a < Z < b) = P(Z < b) - P(Z < a)

Theorems

Standard Normal Distribution
Properties of Normal Distribution

Suitable Grade Level

Undergraduate / Advanced High School