https://file.aiquickdraw.com/mathbot/file/17292376979160x4k2w06.jpg

Let's solve the questions based on the information provided in the image. The normal distribution has:
- Mean ($\mu$) = $1000
- Standard deviation ($\sigma$) = $90

We'll use the z-score formula to calculate each probability:
\[
z = \frac{X - \mu}{\sigma}
\]
where:
- $X$ is the value we are analyzing.
- $\mu$ is the mean.
- $\sigma$ is the standard deviation.

### Part A: Probability that a buyer paid more than $1153
Calculate the z-score for $X = 1153$:
\[
z = \frac{1153 - 1000}{90} = \frac{153}{90} \approx 1.7
\]
Using the z-table or standard normal distribution table, the area to the left of $z = 1.7$ is approximately 0.9554. The probability of paying more than $1153 is:
\[
P(X > 1153) = 1 - P(Z < 1.7) = 1 - 0.9554 = 0.0446
\]

### Part B: Probability that a buyer paid between $787 and $1012
Calculate the z-scores for both values:
1. For $X = 787$:
   \[
   z = \frac{787 - 1000}{90} = \frac{-213}{90} \approx -2.3667
   \]
   The area to the left of $z = -2.37$ is approximately 0.0085.
   
2. For $X = 1012$:
   \[
   z = \frac{1012 - 1000}{90} = \frac{12}{90} \approx 0.1333
   \]
   The area to the left of $z = 0.13$ is approximately 0.5517.

The probability of paying between $787 and $1012 is:
\[
P(787 < X < 1012) = P(Z < 0.13) - P(Z < -2.37) = 0.5517 - 0.0085 = 0.5432
\]

### Part C: What price would the buyer pay to get 5% most expensive HD televisions?
To find the price corresponding to the top 5% of prices, we need the z-score that leaves 5% in the upper tail of the distribution. The z-score for the 95th percentile (top 5%) is approximately:
\[
z \approx 1.645
\]
We use the z-score formula to find the price $X$:
\[
1.645 = \frac{X - 1000}{90}
\]
\[
X - 1000 = 1.645 \times 90
\]
\[
X = 1000 + 148.05 = 1148.05
\]
So, the buyer would pay about $1148.05 to be in the top 5% of prices.

### Explanation for Part A
The z-score represents how many standard deviations a value is away from the mean. For part A, the z-score was calculated as 1.7, meaning $1153 is 1.7 standard deviations above the mean. Using this z-score, we found the corresponding probability of paying more than $1153 to be 0.0446, or about 4.46%.

Do you have any questions or would you like further details?

### 5 Related Questions
1. How do you interpret a z-score in a real-life scenario?
2. What is the significance of the 95th percentile in normal distributions?
3. How do you calculate probabilities for a normal distribution without a z-table?
4. How do standard deviation and mean affect the shape of the normal distribution?
5. What methods can be used to approximate probabilities when z-scores are not available?

### Tip
When dealing with normal distributions, always sketch the curve and mark the z-scores to visualize areas under the curve. This helps in understanding probabilities intuitively.

What is the probability that a buyer paid more than $1153?
What is the probability that a buyer paid between $787 and $1012?
What price would the buyer pay to get 5% the most expensive HD televisions?

Learn how to calculate the probability that a buyer paid more than $1153 or between $787 and $1012 for a particular HD television model, given a normal distribution with a mean of $1000 and a standard deviation of $90. Also, find out what price corresponds to the top 5% of the distribution.

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