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.