A shipping company claims that 95% of packages are delivered on time. A student wants to conduct a simulation
1 answer:
The average number of packages the led to be selected in order to find a package that was not delivered on time is 2.2 option first is correct.
<h3>What is simple random sampling?</h3>With simple random sampling, each component of the population has an equal chance of being selected for the sample.
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We have:
A shipping company claims that 95% of packages are delivered on time.
From the table:
00- 04 is the package not delivered on time.
05-99 - package delivered on time.
The first simulation's third number, 31, 64, 04, is where you can locate an item that wasn't delivered on time.
The second simulation's second number, 34, 02, is the location of a late-delivered delivery.
The third simulation is equal to 96,00 - the second number, which is the location of a late package.
The second number where to find a shipment not delivered on time in the fourth simulation is 31,04
Where to find a box that wasn't delivered on time in the fifth simulation: 98, 01, second number
The average number = (3 + 2 + 2 + 2 + 2)/5
= 11/5
= 2.2
Thus, the average number of packages the led to be selected in order to find a package that was not delivered on time is 2.2 option first is correct.
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Answer:
14.28% of individual adult females have weights between 75 kg and 83 kg.
92.82% of the sample means are between 75 kg and 83 kg.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
.
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
Assume that weights of adult females are normally distributed with a mean of 79 kg and a standard deviation of 22 kg. This means that .
What percentage of individual adult females have weights between 75 kg and 83 kg?
This percentage is the pvalue of Z when subtracted by the pvalue of Z when
. So:
X = 83
has a pvalue of 0.5714.
X = 75
has a pvalue of 0.4286.
This means that 0.5714-0.4286 = 0.1428 = 14.28% of individual adult females have weights between 75 kg and 83 kg.
If samples of 100 adult females are randomly selected and the mean weight is computed for each sample, what percentage of the sample means are between 75 kg and 83 kg?
Now we use the Central Limit THeorem, when . So
X = 83
has a pvalue of 0.9641.
X = 75
has a pvalue of 0.0359.
This means that 0.9641-0.0359 = 0.9282 = 92.82% of the sample means are between 75 kg and 83 kg.