Suppose that the heights of adult women in the United States are normally distributed with a mean of 63.5 inches and a standard deviation of 2.4 inches. Lena
is taller than of the population of U.S. women. How tall (in inches) is Lena? Carry your intermediate computations to at least four decimal places. Round
your answer to one decimal place.

Question

Suppose that the heights of adult women in the United States are normally distributed with a mean of 63.5 inches and a standard deviation of 2.4 inches. Lena
is taller than

of the population of U.S. women. How tall (in inches) is Lena? Carry your intermediate computations to at least four decimal places. Round
your answer to one decimal place.

UpStudy AI · Accepted Answer

1. We are given that the heights $$ X $$ of U.S. women follow a normal distribution with mean $$\mu = 63.5$$ inches and standard deviation $$\sigma = 2.4$$ inches. That is,  
   $$
   X \sim N(63.5, 2.4).
   $$

2. Lena is taller than $$90\%$$ of the population, which means her height corresponds to the $$90^{\text{th}}$$ percentile of the distribution. Let $$X_L$$ be Lena's height.

3. The z-score corresponding to the $$90^{\text{th}}$$ percentile is denoted by $$z_{0.90}$$. From the standard normal table, we have  
   $$
   z_{0.90} \approx 1.2816.
   $$

4. We use the z-score formula to relate $$X_L$$ and $$z_{0.90}$$:  
   $$
   z = \frac{X_L - \mu}{\sigma}.
   $$
   Plug in the values:  
   $$
   1.2816 = \frac{X_L - 63.5}{2.4}.
   $$

5. Solve for $$X_L$$:  
   $$
   X_L - 63.5 = 1.2816 \times 2.4.
   $$
   Compute the product:  
   $$
   1.2816 \times 2.4 \approx 3.0758.
   $$
   Then add the mean:  
   $$
   X_L \approx 63.5 + 3.0758 = 66.5758.
   $$

6. Rounding to one decimal place, Lena's height is approximately  
   $$
   66.6 \text{ inches}.
   $$
Lena is approximately 66.6 inches tall.

Solución de inteligencia artificial de Upstudy

Responder

Solución

Beyond the Answer

preguntas relacionadas

Latest Statistics Questions