Question

multiple choice 4 points find the indicated percentage. the heights of women are normally distributed with a mean of 65 inches and a standard deviation of 2.5 inches. the heights of men are also normal with a mean of 70 inches. approximately what percent of women are taller than a man of average height? 6% 2% 95% 5% 98% multiple choice 4 points identify the value corresponding to the indicated percentile. a tire company finds the lifespan for one brand of its tires is normally distributed with a mean of 47,500 miles and a standard deviation of 3000 miles. if the manufacturer is willing to replace no more than 10% of the tires, what should be the approximate number of miles for a warranty? 52,435 51,340 43,660 42,565

Explanation:

Step1: Recall the z - score formula and percentile concept for normal distribution

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value from the data set, $\mu$ is the mean and $\sigma$ is the standard deviation. For a normal distribution, we can use the standard normal table (z - table) to find the z - score corresponding to a given percentile.

Step2: Solve for problem 16

The mean height of women $\mu = 65$ inches and standard deviation $\sigma=2.5$ inches. The mean height of men is 70 inches. We want to find the percentage of women taller than a man of average height. First, calculate the z - score: $z=\frac{70 - 65}{2.5}=\frac{5}{2.5}=2$. Looking up the value in the standard - normal table for $z = 2$, the cumulative probability $P(Z\leq2)=0.9772$. The percentage of women taller than 70 inches is $1 - 0.9772=0.0228\approx2\%$.

Step3: Solve for problem 17

The mean of the tire lifespan $\mu = 47500$ miles and standard deviation $\sigma = 3000$ miles. The manufacturer is willing to replace no more than 10% of the tires. We need to find the z - score corresponding to the 90th percentile. Looking up in the standard - normal table, the z - score $z$ corresponding to a cumulative probability of 0.9 is approximately 1.28. Using the z - score formula $z=\frac{x-\mu}{\sigma}$, we can solve for $x$: $1.28=\frac{x - 47500}{3000}$. Cross - multiply: $1.28\times3000=x - 47500$. Then $3840=x - 47500$. Add 47500 to both sides: $x=47500 + 3840=51340$.

Answer:

1. 2%
2. 51,340