An average light bulb manufactured by the Acme Corporation lasts 300 days with a standard deviation of 50 days. Assuming that bulb life is normally distributed:
1. What is the probability that an Acme light bulb will last more than 300 days?
2. What is the probability that an Acme light bulb will last less than 300 days?
3. What is the probability that an Acme light bulb will last exactly 300 days?
4. In order to obtain a scientific survey with 95 % confidence level of public opining on something without making more than 3% error in either direction, how much percentage of all American adults should we ask?

1. 90% 2. 10% 3. 50%

Step-by-step explanation:

Standard Deviation (σ) = 50 days

Average/Mean (μ) = 300days

Probability that it would last more than 300 days = P(Bulb>300 days)

We will assume there are 365 days in a year.

P(Bulb>300 days)  implies that the bulb would

Using the normal equation;

z = standard/normal score = (x-μ)/σ where x is the value to be standardized

P(Bulb>300 days) implies x = 365 days

Therefore z = (365-300)/50 = 1.3

Using the normal graph for z=1.3, probability = 90%

2. P(Bulb<300 days) = 1 - P(Bulb>300 days)\

P(Bulb<300 days) = 1 - 0.9

P(Bulb<300 days)  = 10%

3. P(Bulb=300 days) implies z=0 since x=300

Using the normal graph for z=0, probability =50%

l.ln.lln.k.nkl

step-by-step explanation:

all group life insurance is term insurance. actually it is annual renewable term and it is rated on the average age and claims experience of the entire group, which is called "experience rating." remember, this type of insurance has a 31-day grace period and is convertible to whole life upon leaving employment. a contributory group plan requires that both the employer and the employees pay part of the premium. at least 75% of those eligible must participate, so the group is sure to get most of the healthy employees as well as those that are sick.

step-by-step explanation: