Lainey is sliding down a water slide that is 441 feet long at a rate of 49 feet per second. How long will it take Lainey to trav
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Answer:
a) 44.93% probability that there are no surface flaws in an auto's interior
b) 0.03% probability that none of the 10 cars has any surface flaws
c) 0.44% probability that at most 1 car has any surface flaws
Step-by-step explanation:
To solve this question, we need to understand the Poisson and the binomial probability distributions.
Poisson distribution:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
In which
x is the number of sucesses
e = 2.71828 is the Euler number
is the mean in the given interval.
Binomial distribution:
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
Poisson distribution with a mean of 0.08 flaws per square foot of plastic panel. Assume an automobile interior contains 10 square feet of plastic panel.
So
(a) What is the probability that there are no surface flaws in an auto's interior?
Single car, so Poisson distribution. This is P(X = 0).
44.93% probability that there are no surface flaws in an auto's interior
(b) If 10 cars are sold to a rental company, what is the probability that none of the 10 cars has any surface flaws?
For each car, there is a probability of having no surface flaws. 10 cars, so n = 10. This is P(X = 10), binomial, since there are multiple cars and each of them has the same probability of not having a surface defect.
0.03% probability that none of the 10 cars has any surface flaws
(c) If 10 cars are sold to a rental company, what is the probability that at most 1 car has any surface flaws?
At least 9 cars without surface flaws. So
0.44% probability that at most 1 car has any surface flaws
Answer:
The answer is 0.31
Step-by-step explanation:
<u>Old Method New Method .</u>
Mean 73.5625 Mean 75.70588
Standard Error 3.143736 Standard Error 2.923994
Median 72 Median 75
Mode 90 Mode 64
Standard deviation 12.57494 Standard deviation 12.05594
Sample Variance 158.1292 Sample Variance 145.3456
Kurtosis -1.14544 Kurtosis -0.76646
Skewness 0.171025 Skewness 0.091008
Range 39 Range 41
Minimum 55 Minimum 56
Maximum 94 Maximum 97
Sum 1177 Sum 1287
Count 16 Count 17
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: μNew< μOld
Alternative hypothesis: μNew > μOld
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the mean difference between sample means is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
DF = 31
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.
The observed difference in sample means produced a t statistic of - 0.499. We use the t Distribution Calculator to find P(t < - 0.499) = 0.311
Therefore, the P-value in this analysis is 0.311.
Interpret results. Since the P-value (0.311) is greater than the significance level (0.05), we cannot reject the null hypothesis.
From the above test we do have sufficient evidence in the favor of the claim that new method is efficient than the old method.
