Question 1161646: PLEASE HELP ----> Given Car fuel efficiency data: Mean = 34 and Standard Deviation = 2.2
Find:
A) Probability 3 randomly selected cars exceed 37.3
B) Original raw score C, given 85% of all cars have efficiencies exceeding C, (namely P(x>C)=.85)
C) Given new random variable y = 3x - 10 Find P(y>102)
(I did the work and I want to check my answers with one you. please show work in case I get it wrong, thanks!)
Answer by CPhill(2285)   (Show Source):
You can put this solution on YOUR website! ```python
import scipy.stats as stats
import math
mu = 34
sigma = 2.2
# A) P(X > 37.3) for 1 car, then for 3 cars (independently)
z_a = (37.3 - mu) / sigma
p_1 = 1 - stats.norm.cdf(z_a)
p_3 = p_1 ** 3
print(f"A) Single car P(X > 37.3): z = {z_a:.4f}, p = {p_1:.6f}")
print(f" For 3 cars: p^3 = {p_3:.8f}")
# B) P(X > C) = 0.85 => P(X <= C) = 0.15
z_b = stats.norm.ppf(0.15)
C = mu + z_b * sigma
print(f"B) z = {z_b:.4f}, C = {C:.4f}")
# C) y = 3x - 10, P(y > 102)
# y > 102 => 3x - 10 > 102 => 3x > 112 => x > 112/3 = 37.333333...
x_c = 112 / 3
z_c = (x_c - mu) / sigma
p_c = 1 - stats.norm.cdf(z_c)
# Alternatively, mean_y = 3*34 - 10 = 92, std_y = 3*2.2 = 6.6
mean_y = 3 * mu - 10
std_y = 3 * sigma
z_c_alt = (102 - mean_y) / std_y
p_c_alt = 1 - stats.norm.cdf(z_c_alt)
print(f"C) mean_y = {mean_y}, std_y = {std_y}")
print(f" z_y = {z_c_alt:.4f}, P(y > 102) = {p_c:.6f}")
```
```text
A) Single car P(X > 37.3): z = 1.5000, p = 0.066807
For 3 cars: p^3 = 0.00029817
B) z = -1.0364, C = 31.7198
C) mean_y = 92, std_y = 6.6000000000000005
z_y = 1.5152, P(y > 102) = 0.064867
```
Here is the step-by-step breakdown so you can check your work:
**Given:** $\mu = 34$, $\sigma = 2.2$
---
### Part A: Probability that 3 randomly selected cars each exceed 37.3
First, find the probability that a **single** car exceeds 37.3:
1. **$z$-score for a single car:**
$$z = \frac{37.3 - 34}{2.2} = \frac{3.3}{2.2} = 1.50$$
2. **Probability for one car:**
$$P(X > 37.3) = P(Z > 1.50) = 1 - 0.9332 = 0.0668$$
3. **Probability for 3 independent cars:**
$$P(\text{All 3 exceed 37.3}) = (0.0668)^3 \approx \mathbf{0.000298}\quad (\text{or } \approx 0.03\%)$$
*(Note: If the question meant the **sample mean** of 3 cars exceeding 37.3 rather than all 3 individually exceeding it, $SE = \frac{2.2}{\sqrt{3}} \approx 1.2702$, leading to $z = \frac{3.3}{1.2702} = 2.60$, giving $P(\bar{X} > 37.3) = 1 - 0.9953 = \mathbf{0.0047}$.)*
---
### Part B: Find raw score $C$ such that $P(X > C) = 0.85$
1. **Find the left-tail probability:**
If $85\%$ of the area is to the *right* of $C$, then $15\%$ ($0.1500$) lies to the *left* of $C$.
2. **Find the corresponding $z$-score:**
Looking up an area of $0.1500$ in a Standard Normal Table gives:
$$z \approx -1.036$$
3. **Convert $z$ to raw score $C$:**
$$C = \mu + z\sigma = 34 + (-1.036)(2.2) = 34 - 2.28 = \mathbf{31.72}$$
---
### Part C: Find $P(y > 102)$ where $y = 3x - 10$
**Method 1: Transform $y$ back to $x$**
1. **Solve for $x$:**
$$3x - 10 > 102 \implies 3x > 112 \implies x > 37.333$$
2. **Calculate $z$-score:**
$$z = \frac{37.333 - 34}{2.2} = \frac{3.333}{2.2} \approx 1.5152 \quad (\approx 1.52)$$
3. **Find probability:**
$$P(Z > 1.52) = 1 - 0.9357 = \mathbf{0.0643}\quad (\text{using exact } z \approx \mathbf{0.0649})$$
---
**Method 2: Find mean and standard deviation of $y$**
1. **Mean of $y$ ($\mu_y$):**
$$\mu_y = 3\mu - 10 = 3(34) - 10 = 92$$
2. **Standard deviation of $y$ ($\sigma_y$):**
$$\sigma_y = 3\sigma = 3(2.2) = 6.6$$
3. **Calculate $z$-score for $y = 102$:**
$$z = \frac{102 - 92}{6.6} = \frac{10}{6.6} \approx 1.5152 \quad (\approx 1.52)$$
4. **Find probability:**
$$P(Y > 102) = P(Z > 1.52) \approx \mathbf{0.0649}$$
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