Question

1. in the united states, the mean average height of adult women is approximately 65.5 inches, with a standard deviation of 2.5 inches. if the height is normally distributed, what percent of heights are between 65.5 and 70.5 inches.

1. at one college, gpas are normally distributed with a mean of 2.9 and a standard deviation of 0.6. which of the following is the percent of students at the college that have a gpa between 2.3 and 3.5?

e) 68%
f) 99.7%
g) 95%
h) 84%

1. the mean september midday temperature in sacramento is 32°c, with a standard deviation of 2.5°c. how many days in september would you expect the midday temperature to be between 27°c and 39.5°c?
2. the mean gas mileage of cars driven by rosemont high school students is 28 mpg, 68% of students have gas mileage between 22 mpg and 34 mpg. what is the standard deviation?

challenge: extra credit question
a machine produces electrical components. 95% have lengths between 1.170 and 1.228 cm. assuming the data is normally distributed, what are the mean and standard deviation?
mean=
standard deviation=

Explanation:

Step1: Calculate z - scores

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 the first problem (height of adult women): $\mu = 65.5$, $\sigma=2.5$, $x_1 = 65.5$ and $x_2 = 70.5$.
The z - score for $x_1$: $z_1=\frac{65.5 - 65.5}{2.5}=0$.
The z - score for $x_2$: $z_2=\frac{70.5 - 65.5}{2.5}=\frac{5}{2.5} = 2$.

Step2: Use the standard normal distribution table

The area under the standard - normal curve corresponding to $z = 0$ is $0.5$ and the area corresponding to $z = 2$ is $0.9772$.
The area between $z = 0$ and $z = 2$ is $0.9772-0.5=0.4772$.
So the percentage of heights between 65.5 and 70.5 inches is $47.72\%$.

For the second problem (GPA): $\mu = 2.9$, $\sigma = 0.6$, $x_1 = 2.3$ and $x_2 = 3.5$.
The z - score for $x_1$: $z_1=\frac{2.3 - 2.9}{0.6}=\frac{- 0.6}{0.6}=-1$.
The z - score for $x_2$: $z_2=\frac{3.5 - 2.9}{0.6}=\frac{0.6}{0.6}=1$.
The area corresponding to $z=-1$ is $0.1587$ and the area corresponding to $z = 1$ is $0.8413$.
The area between $z=-1$ and $z = 1$ is $0.8413 - 0.1587=0.6826\approx68\%$. So the answer is e) 68%.

For the third problem (temperature in Sacramento): $\mu = 32$, $\sigma = 2.5$, $x_1 = 27$ and $x_2 = 39.5$.
The z - score for $x_1$: $z_1=\frac{27 - 32}{2.5}=\frac{-5}{2.5}=-2$.
The z - score for $x_2$: $z_2=\frac{39.5 - 32}{2.5}=\frac{7.5}{2.5}=3$.
The area corresponding to $z=-2$ is $0.0228$ and the area corresponding to $z = 3$ is $0.9987$.
The area between $z=-2$ and $z = 3$ is $0.9987-0.0228 = 0.9759$.
September has 30 days. So the expected number of days is $30\times0.9759\approx29$ days.

For the fourth problem (gas mileage): Since 68% of the data lies within one standard deviation of the mean in a normal distribution, and $\mu = 28$, $x_1 = 22$, $x_2 = 34$.
We know that $\mu-\sigma=22$ and $\mu+\sigma=34$. Substituting $\mu = 28$ into $\mu-\sigma=22$, we get $28-\sigma=22$, so $\sigma = 6$.

For the extra - credit problem: Since 95% of the data lies within two standard deviations of the mean in a normal distribution, if $x_1 = 1.170$ and $x_2 = 1.228$.
The mean $\mu=\frac{1.170 + 1.228}{2}=\frac{2.398}{2}=1.199$.
Since $x_2=\mu + 2\sigma$ and $\mu = 1.199$, $x_2 = 1.228$. Then $1.228=1.199 + 2\sigma$.
$2\sigma=1.228 - 1.199=0.029$, so $\sigma=\frac{0.029}{2}=0.0145$.

Answer:

1. $47.72\%$
2. e) 68%
3. 29 days
4. 6
5. Mean: $1.199$, Standard Deviation: $0.0145$