Answer:
8% or 0.08
Step-by-step explanation:
Probability of missing the first pass = 40% = 0.40
Probability of missing the second pass = 20% = 0.20
We have to find the probability that he misses both the passes. Since the two passes are independent of each other, the probability that he misses two passes will be:
Probability of missing 1st pass x Probability of missing 2nd pass
i.e.
Probability of missing two passes in a row = 0.40 x 0.20 = 0.08 = 8%
Thus, there is 8% probability that he misses two passes in a row.
Answer:
3 = 95, 4 = 85, if a and b are parallel
Step-by-step explanation:
Step-by-step explanation:
Mean= Sumof all the numbers/ number of data
So,
23+37+11+58+13+45=187
Now...
187/6=31.166666667 Answer
hope it helps....
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.