Answer:
P(pumps) = 2/5
Step-by-step explanation:
(Pumps)/(Everything)
(10)/(9+6+10)
10/25
2/5
The question is incomplete! Complete question along with answer and step by step explanation is provided below.
Question:
The lifetime (in hours) of a 60-watt light bulb is a random variable that has a Normal distribution with σ = 30 hours. A random sample of 25 bulbs put on test produced a sample mean lifetime of = 1038 hours.
If in the study of the lifetime of 60-watt light bulbs it was desired to have a margin of error no larger than 6 hours with 99% confidence, how many randomly selected 60-watt light bulbs should be tested to achieve this result?
Given Information:
standard deviation = σ = 30 hours
confidence level = 99%
Margin of error = 6 hours
Required Information:
sample size = n = ?
Answer:
sample size = n ≈ 165
Step-by-step explanation:
We know that margin of error is given by
Margin of error = z*(σ/√n)
Where z is the corresponding confidence level score, σ is the standard deviation and n is the sample size
√n = z*σ/Margin of error
squaring both sides
n = (z*σ/Margin of error)²
For 99% confidence level the z-score is 2.576
n = (2.576*30/6)²
n = 164.73
since number of bulbs cannot be in fraction so rounding off yields
n ≈ 165
Therefore, a sample size of 165 bulbs is needed to ensure a margin of error not greater than 6 hours.
Answer:
37%
Step-by-step explanation:
If Ramona received an overall pay of $52,561 last year, $23,960 of which was base pay, we can first subtract these two numbers to find the amount she made from her sales commission:
$52,561 - $23,960 = $28,601
In order to find her percent commission, we can set up a proportion:
, where 'p' is her rate of commission
Cross-multiply and divide: 100(28601) = 77300x or 2,860,100 = 77,300x
Or x = 37%
