Question

1. what is the mean and standard deviation of this normal distribution?

a. mean = 1010; sd = 60
b. mean = 20; sd = 1010
c. mean = 1000; sd = 30
d. mean = 1010; sd = 20

1. a machine produces electrical components. 95% of the components have lengths between 1.75 cm

1.95 cm. assuming this data is normally distributed, what are the mean and standard deviation?
a. mean = 1.80; sd = 0.05
b. mean = 1.85; sd = 0.05
c. mean = 1.80; sd = 0.50
d. mean = 1.85; sd = 0.50

1. the average age of a group of fans at a hockey game is 36 years with a standard deviation of 8

years. what percent of fans are between 20 and 28 years old?
a. 34%
b. 16%
c. 13.5%
d. 2.5%

Explanation:

Step1: Recall the properties of normal distribution

In a normal distribution, the mean is at the center of the bell - shaped curve. For question 4, the center of the given normal distribution curve is at \(x = 1010\). The standard deviation can be calculated using the distance between the mean and the inflection points. The distance between \(1010\) and \(990\) (or \(1030\) and \(1010\)) is \(20\).

Step2: Use the empirical rule for question 5

For a normal distribution, about \(95\%\) of the data lies within \(\mu\pm2\sigma\). If \(95\%\) of the components have lengths between \(1.75\) cm and \(1.95\) cm, then \(\mu=\frac{1.75 + 1.95}{2}=1.85\) and \(2\sigma=1.95 - 1.75=0.2\), so \(\sigma = 0.1\div2=0.05\)

Step3: Use the z - score formula for question 6

The z - score formula is \(z=\frac{x-\mu}{\sigma}\). For \(x = 20\), \(z_1=\frac{20 - 36}{8}=\frac{- 16}{8}=-2\). For \(x = 28\), \(z_2=\frac{28 - 36}{8}=\frac{-8}{8}=-1\). Using the standard normal distribution table, the area to the left of \(z=-1\) is \(0.1587\) and the area to the left of \(z = - 2\) is \(0.0228\). The area between \(z=-2\) and \(z=-1\) is \(0.1587-0.0228 = 0.1359\approx13.5\%\)

Answer:

1. D. Mean \(=1010\); SD \(=20\)
2. B. Mean \(=1.85\); SD \(=0.05\)
3. C. \(13.5\%\)