STAT200/STAT 200 Week 4 Homework Problems (summer 2020) complete solutions.

STAT 200 Week 4 Homework Problems
6.1.2

1.) The commuter trains on the Red Line for the Regional Transit Authority (RTA) in Cleveland, OH,
have a waiting time during peak rush hour periods of eight minutes ("2012 annual report," 2012).
a.) State the random variable.
a. The mean waiting time during peak rush hours
b.) Find the height of this uniform distribution.
1 1
a. 𝐻𝑒𝑖𝑔ℎ𝑡 = 8−0 = 8 = 0.125
c.) Find the probability of waiting between four and five minutes.
a. 𝑃(4 < 𝑥 < 5) = 𝑃(𝑥 < 5) − 𝑃(𝑥 < 4)
1 1
=5x –4x
8 8
= (5 x 0.125) – (4 x 0.125)
= 0.125
d.) Find the probability of waiting between three and eight minutes.
a. 𝑃(3 < 𝑥 < 8) = 8(0.125) – 3(0.125)
= 1 – 0.375
= 0.625
e.) Find the probability of waiting five minutes exactly.
a. 𝑃(𝑥 = 5) = 0 x 0.125
=0


6.3.2

Find the z-score corresponding to the given area. Remember, z is distributed as the standard normal
distribution with mean of m = 0 and standard deviation s = 1.
a.) The area to the left of z is 15%.
a. z-score = -1.0364
b.) The area to the right of z is 65%.
a. Z-score = -0.3853
c.) The area to the left of z is 10%.
a. Z-score = -1.2861
d.) The area to the right of z is 5%.
a. Z-score = 1.6449
e.) The area between -z and z is 95%. (Hint draw a picture and figure out the area to the left
of the -z .)
a. Z-score = ±1.96
f.) The area between -z and z is 99%.
a. Z-score = ±2.5758
6.3.4

According to the WHO MONICA Project the mean blood pressure for people in China is 128 mmHg with
a standard deviation of 23 mmHg (Kuulasmaa, Hense & Tolonen, 1998). Assume that blood pressure is
normally distributed.

a.) State the random variable.

, a. The mean blood pressure for people in China
b.) Find the probability that a person in China has blood pressure of 135 mmHg or more.
𝑥−𝜇 135−128
a. 𝑃(𝑥 ≥ 135) = 𝑃 ( 𝜎
)
= 𝑃( 23 )
= 𝑃(𝑧 ≥ 0.3043)
= 0.3804
c.) Find the probability that a person in China has blood pressure of 141 mmHg or less.
𝑥−𝜇 141−128
a. 𝑃(𝑥 ≤ 141) = 𝑃 ( 𝜎
)
= 𝑃( 23 )
= 𝑃(𝑧 ≤ 0.565)
= 0.714
d.) Find the probability that a person in China has blood pressure between 120 and 125 mmHg.
𝑥−𝜇 141−128
a. 𝑃(𝑥 ≤ 141) = 𝑃 ( 𝜎
)
= 𝑃( 23 )
= 𝑃(𝑧 ≤ 0.565)
= 0.714
e.) Is it unusual for a person in China to have a blood pressure of 135 mmHg? Why or why not?
a. No, it is not unusual for a person in China to have a blood pressure of 135 mmHg,
since the probability of having this blood pressure reading is not less than 5%.
f.) What blood pressure do 90% of all people in China have less than?
a. Blood pressure = α
𝑃 (𝑥 ≤ 𝑎) = 0.9
𝑥−𝜇 𝑎−𝜇
𝑃( 𝜎
) = 𝑃( 𝜎
)= 0.9

From the standard normal table, 𝑃(𝑧 < 1.285) = 0.9
𝜇−128
= 1.285
23

𝜇 = 1.285(23) + 123
= 157.55 is the blood pressure

6.3.8

A dishwasher has a mean life of 12 years with an estimated standard deviation of 1.25 years ("Appliance
life expectancy," 2013). Assume the life of a dishwasher is normally distributed.

a.) State the random variable.
a. The mean dishwasher life
b.) Find the probability that a dishwasher will last more than 15 years.
𝑥−𝜇 15−12
a. 𝑃(𝑥 ≥ 15) = 𝑃 ( ) =
𝜎 1.25
= 𝑃(𝑧 > 2.4)
= 0.01255
c.) Find the probability that a dishwasher will last less than 6 years.
𝑥−𝜇 15−12
a. 𝑃(𝑥 ≤ 6) = 𝑃 ( 𝜎 ) = 1.25
= 𝑃(𝑧 > 2.4)
= 0.01255
d.) Find the probability that a dishwasher will last between 8 and 10 years.