##### General Statistics: Ch 6 Quiz

College: First year, College: Second year, College: Third year, College: Fourth year

Assume that thermometer readings are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. A thermometer is randomly selected and tested.

For the case below, draw a sketch, and find the probability of the reading. (The given values are in Celsius degrees.)

Between –1.47 and 1.87

The probability of getting a reading between –1.47°C and 1.87°C is **
0.8985**
** **.

*P(-1.47 < z < 1.87)*

*= (0.9693 – 0.0708)*

Assume the readings on thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C.

Find the probability that a randomly selected thermometer reads greater than 0.04 and draw a sketch of the region.

The probability is **0.4840**.

*P(z > 0.04)*

*= 1 – P(z < 0.04)*

*= 1 – 0.5160 = 0.4840*

*OR*

*Symmetric to z score of -0.04 (Area = 0.4840)*

Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1.

Find the probability that a given score is less than –0.56 and draw a sketch of the region.

The probability is ** 0.2877**.

P(z < -0.56) = 0.2877

Find the indicated z score. The graph depicts the standard normal distribution with mean 0 and standard deviation 1.

The indicated z score is ** 2.09**.

Find the indicated z score. The graph depicts the standard normal distribution with mean 0 and standard deviation 1.

The indicated z score is ** –0.54**.

*Symmetric to Area of 0.7054 (z = 0.54)*

*OR*

*Area = 1 – 0.7054 = 0.2946*

A statistics professor plans classes so carefully that the lengths of her classes are uniformly distributed between 49.0 and 59.0 minutes.

Find the probability that a given class period runs between 50.5 and 51.5 minutes.

**0.**
**1** **00**

* * * 49.0 = 10.0*

* Since the uniform distribution is rectangular,
has a length of 10.0, and an area of 1,
the height of the uniform distribution is 0.1.*

* P(between 50.5 and 51.5) = (shaded region length) × (shaded
region height)*

* = (51.5 – 50.5) × (0.1)*

* = 0.1*

Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1.

Find the probability that a given score is less than 3.69 and draw a sketch of the region.

The probability is ** 0.9999**.

Find the area of the shaded region. The graph depicts the standard normal distribution of bone density scores with mean 0 and standard deviation 1.

The area of the shaded region is ** 0.8508**.

*P(z > 0.04)*

*= 1 – P(z < -1.04)*

*= 1 – 0.1492 = 0.8505*

*OR*

*Symmetric to z score of 1.04 (Area = 0.8508)*

The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 0 and 6 minutes. Find the probability that a randomly selected passenger has a waiting time greater than 4.25 minutes.

**0.298**

*P(greater than 45.25)
= (length of shaded region) × (height of shaded region)*

*= (6 – 5.25) × (0.167)-round to 2 dec. places*

*= (1.75 × (0.17)*

* = 0.2975*

Assume that thermometer readings are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. A thermometer is randomly selected and tested.

For the case below, draw a sketch, and find the probability of the reading. (The given values are in Celsius degrees.)

Between –0.50 and 2.00

The probability of getting a reading between 0.50°C and 2.00°C is **
0.2857**.

*P(0.50 < z < 2.00)*

*= (0.9772 – 0.6915)*