Question

1.-

A company produces scooters used by small businesses that find them convenient for making short deliveries. The company is notified whenever a scooter breaks​ down, and the problem is classified as being either mechanical or electrical. The company then matches the scooter to the plant where it was assembled. The data table to the right contains a random sample of 200 breakdowns. Use the data file and the relative frequency assessment method to complete parts a through c.			Electrical	Mechanical	Total
		Plant 1	28	43	71
		Plant 2	68	61	129
		Total	96	104	200

a. What is the probability a scooter was assembled at Plant 1​?

​P(Plant 1​)=

​(Type an integer or a​ decimal.)

b. What is the probability that a scooter breakdown was due to a mechanical ​problem?

​P(mechanical​)=

​(Type an integer or a​ decimal.)

c. What is the probability that a scooter breakdown was due to an electrical problem and the scooter was assembled at Plant 2​?

​P(electrical and Plant 2​)=

​(Type an integer or a​ decimal.)

2.-

A population is normally distributed with

μ=100

and

σ=10.

a.	Find the probability that a value randomly selected from this population will have a value greater than 115.
b.	Find the probability that a value randomly selected from this population will have a value less than 90.
c.	Find the probability that a value randomly selected from this population will have a value between 90 and 115.

a. ​P(x>115​)=

​(Round to four decimal places as​ needed.)

b.​P(x<90​)=

​(Round to four decimal places as​ needed.)

c. ​P(90<x<115​)=

​(Round to four decimal places as​ needed.)

Answer 1

1) We have been given the below data

	Electrical	Mechanical	Total
Plant 1	28	43	71
Plant 2	68	61	129
Total	96	104	200

The relative frequency for this data is as below

	Electrical	Mechanical	Total
Plant 1	0.14	0.215	0.355
Plant 2	0.34	0.305	0.645
Total	0.48	0.52	1

Relative frequency is calculated by dividing the number of outcomes by the total number of outcomes. For example relative frequency for electrical failures in scooters from Plant 1 is 28/200 = 0.14

a) What is the probability a scooter was assembled at Plant 1​?

We can see that the relative frequency for Plant 1 is 0.355. Hence the probability that the scooter was assembled at Plant 1, P(Plant 1) is 0.355

b) What is the probability that a scooter breakdown was due to a mechanical ​problem?

We can see that the relative frequency for mechnical problem is 0.52. Hence the probability that the scooter breakdown was due to mechanical problem, P(mechanical) is 0.52

c) What is the probability that a scooter breakdown was due to an electrical problem and the scooter was assembled at Plant 2​?

We can see that the relative frequency for electrical problem and Plant 2 is 0.34. Hence the probability that the scooter breakdown was due to electrical problem and the scooter was from Plant 2, P(electrical and Plant 2) is 0.34

2) We have been given that the population is normally distributed with μ=100 and σ=10

a) P(x>115​)= 1-P(x<115)

P(x < 115) = P(z < 115 – 4) = P(z< 115 – 100 -) = P(Z < 1.5) 10

From normal tables, we can see that P(Z<1.5) = 0.9332

Hence P(x>115) = 1-P(x<115) = 1-P(z<1.5) = 1-0.9332 = 0.0668

$b) P(x<90) = P(z < \frac{90-\mu}{\sigma}) = P(z<\frac{90-100}{10}) = P(z<-1)$

From normal tables, we can see that P(Z<-1) = 0.15867

Hence P(x<90) = 0.1587

c) P(90<x<115)

We have found from part a and b that P(x>115) = 0.0668 and P(x<90) = 0.1587. Hence we have the below

P(90<x<115) = 1-P(x<90)-P(x>115) = 1-0.0668-0.1587 = 0.7745