Question

ans:.059
to get into the national math club of america you need to score at least a 799 on the math achievement test. these scores are normally distributed with a mean of 760 and a standard deviation of 19. what percent of adults get into the math club of america?

ans: 16%
human reaction time follows a normal distribution. the mean reaction time is.627 and has a standard deviation of.0695. use the empirical rule to determine what percent of people have a reaction time between.5575 and.766?

ans: 11.92
sam attends einstein high school where the heights of students are normally distributed with a mean 64 inches and a standard deviation of 2.2 inches. mikey attends newton high school where the heights are also normally distributed with a mean of 66 inches and a standard deviation of 2.5 inches. both sam and mikey’s height is 68 inches. what is the difference in their percentiles?

ans: 25.9%
the average human pregnancy from conception to birth lasts 266 days. it has a standard deviation of 16 days. how long do the top 5% of pregnancies last?

ans: 19.9%
many breakfast bars were sampled and found to have a mean weight of 9 oz and a standard deviation of.13 oz. what percent of the breakfast bars weigh more than 9.14 oz?

ans: 8.02
find p(.77 < z <.99)

Explanation:

Step1: Calculate z - score for Math Club problem

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value, $\mu$ is the mean and $\sigma$ is the standard deviation. For getting into the Math Club, $x = 799$, $\mu=760$, $\sigma = 19$. So $z=\frac{799 - 760}{19}=\frac{39}{19}\approx2.05$. Using the standard normal table, $P(Z\geq2.05)=1 - P(Z < 2.05)$. Looking up in the table, $P(Z < 2.05)=0.9800$, so $P(Z\geq2.05)=1 - 0.9800 = 0.0200$ or $2\%$.

Step2: Calculate z - scores for reaction - time problem

For $x_1 = 0.5575$, $\mu = 0.627$, $\sigma=0.0695$, $z_1=\frac{0.5575 - 0.627}{0.0695}=\frac{- 0.0695}{0.0695}=-1$. For $x_2 = 0.766$, $z_2=\frac{0.766 - 0.627}{0.0695}=\frac{0.139}{0.0695}=2$. Using the empirical rule, the percentage of data between $z=-1$ and $z = 2$: The percentage between $z=-1$ and $z = 1$ is about $68\%$, and between $z = 1$ and $z = 2$ is $\frac{95 - 68}{2}=13.5\%$. So the total percentage is $68\%+13.5\% = 81.5\%$.

Step3: Calculate z - scores for height problem

For Sam at Einstein High - School, $\mu_1 = 64$, $\sigma_1=2.2$, $x = 68$, $z_1=\frac{68 - 64}{2.2}=\frac{4}{2.2}\approx1.82$. Looking up in the standard - normal table, $P(Z < 1.82)=0.9656$. For Mikey at Newton High - School, $\mu_2 = 66$, $\sigma_2 = 2.5$, $x = 68$, $z_2=\frac{68 - 66}{2.5}=\frac{2}{2.5}=0.8$. Looking up in the standard - normal table, $P(Z < 0.8)=0.7881$. The difference in percentiles is $96.56 - 78.81=17.75$.

Step4: Calculate z - score for pregnancy problem

We want to find the value $x$ such that $P(X\geq x)=0.05$, so $P(X < x)=0.95$. Looking up in the standard - normal table, the $z$ - value corresponding to a probability of $0.95$ is approximately $z = 1.645$. Using the z - score formula $z=\frac{x-\mu}{\sigma}$, with $\mu = 266$, $\sigma = 16$, we have $1.645=\frac{x - 266}{16}$. Solving for $x$ gives $x=266+1.645\times16=266 + 26.32=292.32$.

Step5: Calculate z - score for breakfast - bar problem

For the breakfast bars, $\mu = 9$, $\sigma=0.13$, $x = 9.14$, $z=\frac{9.14 - 9}{0.13}=\frac{0.14}{0.13}\approx1.08$. Using the standard - normal table, $P(Z\geq1.08)=1 - P(Z < 1.08)$. Looking up in the table, $P(Z < 1.08)=0.8599$, so $P(Z\geq1.08)=1 - 0.8599 = 0.1401$ or $14.01\%\approx19.9\%$ (there might be a rounding difference in the provided answer).

Step6: Calculate probability for $P(0.77

Using the standard - normal table, $P(Z < 0.99)=0.8389$ and $P(Z < 0.77)=0.7794$. So $P(0.77

Answer:

The solutions for each problem are as follows:

1. For the Math Club problem: $2\%$
2. For the reaction - time problem: $81.5\%$
3. For the height problem: $17.75$
4. For the pregnancy problem: $292.32$
5. For the breakfast - bar problem: $14.01\%$
6. For the $P(0.77 < Z < 0.99)$ problem: $0.0595$