A. Consider a normal distribution with mean 31 and standard deviation 6. What is the probability a value selected at random from this distribution is greater than 31? (Round your answer to two decimal places.)
B. Suppose 85% of the area under the standard normal curve lies to the right of z. Is z positive or negative?
C. Assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probability. (Round your answer to four decimal places.)
µ = 4; s = 2
P(3 = x = 7) =
D. Assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probability. (Round your answer to four decimal places.)
µ = 40; s = 14
P(50 = x = 70) =
The Correct Answer and Explanation is:
Let’s go through each part of the question one by one, providing correct answers and detailed explanations.
A. Probability that a value is greater than the mean in a normal distribution
- Given: Mean (μ) = 31, Standard deviation (σ) = 6
- We want to find: $$
P(X > 31)
$$
Since the distribution is normal and symmetric, the probability that a value is greater than the mean is always 0.50.
✅ Answer: 0.50
Explanation:
In a normal distribution, the mean (μ) divides the distribution into two equal halves. Because the distribution is symmetric about the mean, 50% of the data lies to the left of the mean and 50% lies to the right. Therefore, the probability that a randomly selected value from the distribution is greater than 31 (i.e., greater than the mean) is:
$$
P(X > μ) = 0.50
$$
B. If 85% of the area under the standard normal curve lies to the right of z, is z positive or negative?
✅ Answer: Negative
Explanation:
In the standard normal distribution (mean = 0), the right-hand tail gets smaller as z increases. If 85% of the area lies to the right, that means only 15% lies to the left. A high percentage of area to the right implies a z-value located to the left of the mean, because most of the distribution is after it. This can only occur when the z-score is negative.
Mathematically, if $P(Z > z) = 0.85$, then $z \approx -1.036$. So z is negative.
C. Find P(3 ≤ x ≤ 7) for µ = 4, σ = 2
We use the standardization formula:
$$
z = \frac{x – \mu}{\sigma}
$$
Convert x-values to z-scores:
- For x = 3: $z = \frac{3 – 4}{2} = -0.5$
- For x = 7: $z = \frac{7 – 4}{2} = 1.5$
Now, find area between z = -0.5 and z = 1.5 using a standard normal table or calculator:
- $P(Z < 1.5) = 0.9332$
- $P(Z < -0.5) = 0.3085$
So,
$$
P(3 \le x \le 7) = 0.9332 – 0.3085 = 0.6247
$$
✅ Answer: 0.6247
D. Find P(50 ≤ x ≤ 70) for µ = 40, σ = 14
Convert to z-scores:
- For x = 50: $z = \frac{50 – 40}{14} = \frac{10}{14} \approx 0.7143$
- For x = 70: $z = \frac{70 – 40}{14} = \frac{30}{14} \approx 2.1429$
Use a Z-table or calculator:
- $P(Z < 2.14) \approx 0.9834$
- $P(Z < 0.71) \approx 0.7611$
So,
$$
P(50 \le x \le 70) = 0.9834 – 0.7611 = 0.2223
$$
✅ Answer: 0.2223
Summary of Answers:
- A: 0.50
- B: Negative
- C: 0.6247
- D: 0.2223
These calculations are based on the properties of the normal distribution, which is defined by its mean and standard deviation. Converting raw scores (x-values) into z-scores allows us to use standard tables or statistical software to compute exact probabilities
.