Math Problem Statement

We are given a normally distributed random variable X with µ = 75 and σ = 5.

Find P (X > 85).

Find k such as P(X<k) = 0.997. Interpret your result.

Find the two values where the middle 68.27% of the distribution of X lie.

Solution

We are given that the random variable XX is normally distributed with μ=75\mu = 75 and σ=5\sigma = 5, meaning:

XN(75,52)X \sim N(75, 5^2)

1. Find P(X>85)P(X > 85)

We first standardize the value of 85 by converting it to the standard normal form ZZ, where: Z=XμσZ = \frac{X - \mu}{\sigma}

So, for X=85X = 85: Z=85755=105=2Z = \frac{85 - 75}{5} = \frac{10}{5} = 2

Now, we need to find P(X>85)=P(Z>2)P(X > 85) = P(Z > 2).

Using standard normal tables or a calculator: P(Z>2)=1P(Z2)10.9772=0.0228P(Z > 2) = 1 - P(Z \leq 2) \approx 1 - 0.9772 = 0.0228

Thus: P(X>85)0.0228P(X > 85) \approx 0.0228

2. Find kk such that P(X<k)=0.997P(X < k) = 0.997

We need to find the value of kk such that P(X<k)=0.997P(X < k) = 0.997. This means finding the corresponding ZZ-value for which the cumulative probability is 0.997.

Using standard normal tables, we find that: P(Z<2.747)0.997P(Z < 2.747) \approx 0.997

Now, convert the ZZ-value back to XX using: k=Zσ+μk = Z \cdot \sigma + \mu

So: k=2.7475+75=13.735+75=88.735k = 2.747 \cdot 5 + 75 = 13.735 + 75 = 88.735

Thus, the value of kk is approximately 88.74. This means that 99.7% of the distribution of XX lies below 88.74.

3. Find the two values where the middle 68.27% of the distribution lies

The middle 68.27% of the distribution corresponds to one standard deviation above and below the mean in a normal distribution.

For the standard normal distribution, this means: μσtoμ+σ\mu - \sigma \quad \text{to} \quad \mu + \sigma

Thus: 755=70and75+5=8075 - 5 = 70 \quad \text{and} \quad 75 + 5 = 80

So, the middle 68.27% of the distribution lies between 70 and 80.

Summary of Results:

  1. P(X>85)0.0228P(X > 85) \approx 0.0228
  2. k88.74k \approx 88.74 such that P(X<k)=0.997P(X < k) = 0.997
  3. The middle 68.27% of the distribution lies between 70 and 80.

Would you like more details or clarification on any part of this?

Here are some related questions to expand on this topic:

  1. How do you compute probabilities for other values of XX?
  2. What is the 95th percentile for this distribution?
  3. How do changes in μ\mu and σ\sigma affect the distribution?
  4. How can you interpret the result for P(X>85)P(X > 85) in practical terms?
  5. What is the probability that XX is between 80 and 85?

Tip: When dealing with normal distributions, always standardize using the formula Z=XμσZ = \frac{X - \mu}{\sigma} to simplify the calculations!

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Math Problem Analysis

Mathematical Concepts

Statistics
Probability
Normal Distribution
Standard Normal Distribution

Formulas

Z = (X - µ) / σ
P(X > 85) = P(Z > 2)
k = Z * σ + µ

Theorems

68-95-99.7 Rule (Empirical Rule)

Suitable Grade Level

College-level Statistics

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