The prices paid for a particular model of OLED television is normally distributed with a mean price of $1500 and a standard deviation of $95. 


Use this information to fill in the blanks and answer the questions. 
Round probabilities to four decimal places.


a. What is the probability that a randomly selected buyer paid more than $1700?

             

b. What is the probability that a randomly selected buyer paid between $1400 and $1600?

            

c. What is the probability that a randomly selected buyer paid less than $1350?

The problem states that the prices paid for a particular OLED television are normally distributed with a mean of $1500 and a standard deviation of $95. We will use the standard normal distribution (Z-distribution) to calculate the probabilities for each of the scenarios. For this, we will convert the prices into Z-scores using the formula:

\[
Z = \frac{X - \mu}{\sigma}
\]

Where:
- $X$ is the price,
- $\mu = 1500$ (mean),
- $\sigma = 95$ (standard deviation).

### (a) Probability that a randomly selected buyer paid more than $1700:

1. First, calculate the Z-score for $X = 1700$:

\[
Z = \frac{1700 - 1500}{95} = \frac{200}{95} \approx 2.1053
\]

2. Using a standard normal distribution table or a calculator, find the probability that $Z > 2.1053$. The area to the left of $Z = 2.1053$ is approximately 0.9824. Therefore, the probability of paying **more** than $1700 is:

\[
P(X > 1700) = 1 - 0.9824 = 0.0176
\]

So, the probability is approximately **0.0176**.

---

### (b) Probability that a randomly selected buyer paid between $1400 and $1600:

1. Calculate the Z-score for $X = 1400$:

\[
Z = \frac{1400 - 1500}{95} = \frac{-100}{95} \approx -1.0526
\]

2. Calculate the Z-score for $X = 1600$:

\[
Z = \frac{1600 - 1500}{95} = \frac{100}{95} \approx 1.0526
\]

3. Find the probabilities corresponding to these Z-scores:
   - For $Z = -1.0526$, the area to the left is approximately **0.1462**.
   - For $Z = 1.0526$, the area to the left is approximately **0.8538**.

4. The probability that a buyer paid between $1400 and $1600 is the difference between these two probabilities:

\[
P(1400 < X < 1600) = 0.8538 - 0.1462 = 0.7076
\]

So, the probability is approximately **0.7076**.

---

### (c) Probability that a randomly selected buyer paid less than $1350:

1. Calculate the Z-score for $X = 1350$:

\[
Z = \frac{1350 - 1500}{95} = \frac{-150}{95} \approx -1.5789
\]

2. Using a standard normal distribution table or a calculator, find the probability corresponding to $Z = -1.5789$. The area to the left of this Z-score is approximately **0.0572**.

Therefore, the probability that a buyer paid less than $1350 is approximately **0.0572**.

---

### Summary of Answers:
- (a) The probability that a randomly selected buyer paid more than $1700 is **0.0176**.
- (b) The probability that a randomly selected buyer paid between $1400 and $1600 is **0.7076**.
- (c) The probability that a randomly selected buyer paid less than $1350 is **0.0572**.

---

Would you like more details or have any questions on this?

### Related Questions:
1. What is the probability that a buyer paid between $1300 and $1700?
2. What is the price at which 90% of the buyers paid more?
3. What is the probability that a buyer paid exactly the mean price?
4. If you were to randomly select 10 buyers, what is the probability that exactly one of them paid more than $1700?
5. What price would correspond to the 95th percentile of the distribution?

### Tip:
In normal distribution problems, converting raw scores to Z-scores allows you to use standard tables or calculators for probabilities, simplifying the process!

The prices paid for a particular model of OLED television is normally distributed with a mean price of $1500 and a standard deviation of $95. What is the probability that a randomly selected buyer paid more than $1700, between $1400 and $1600, or less than $1350?

Explore probability calculations for OLED TV prices, normally distributed with a mean of $1500 and a standard deviation of $95. Learn how to find the probabilities for prices above $1700, between $1400 and $1600, and below $1350 using Z-scores and standard normal distribution.

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