Question

AB automobile battery is known to last an average of 1600 days with a standard deviation of 99 day. If 324 of these batteries are selected, find the following probabilities for the average length of life of the selected batteries:

a. The average is between 1580 and 1600.

b. The average is greater than 1610.

c. The average is less than 1585.

Rounded to 4 decimal places

Question 14 options:

	(a) 0.5001 (b) 0.0355 (c) 0.0032
	(a) 0.5010 (b) 0.0345 (c) 0.0032
	(a) 0.0800 (b) 0.4598 (c) 0.4398
	(a) 0.4999 (b) 0.0345 (c) 0.0032
	None of these options
	(a) 0.4999 (b) 0.0345 (c) 0.0000

Answer 1

Correct option:

(a) 0.4999 (b) 0.0345 (c) 0.0032

Explanation:

(a)

= 1600

= 99

n = 324

SE = /

= 99/

= 5.50

To find P(1580< < 1600):

Z = 1580 - 1600)/5.50

= - 3.6364

By Technology, Cumulative Area Under Standard Normal Curve = 0.0001

So

P(1580< < 1600):= 0.5000 - 0..0001 = 0.4999

(b)

= 1600

= 99

n = 324

SE = /

= 99/

= 5.50

To find P( > 1610):

Z = (1610 - 1600)/5.50

= 1.8182

By Technology, Cumulative Area Under Standard Normal Curve = 0.9655

So

P( > 1610):= 1 - 0.9655 = 0.0345

(c)

= 1600

= 99

n = 324

SE = /

= 99/

= 5.50

To find P( < 1585):

Z = (1585 - 1600)/5.50

= - 2.7273

By Technology, Cumulative Area Under Standard Normal Curve = 0.0032

So

P( < 1585):= 0.0032