The sample data below shows the number of hours spent by five students over the weekend to prepare for Monday's Business Statist
ics exam. 3 12 2 3 5(See the Excel Data File.) The 75 th percentile of the data is the closest to ________.g
1 answer:
Answer:
8.5 hrs
Step-by-step explanation:
-A 75th percentile mathematically means that 75% of the time data points are below that value and 25% of the time are above that value.
-We plot the given our data {3 12 2 3 5} to determine the 75th percentile
#We can use to plot our boxplot.
-From the plot, we find our 75th percentile to be 8.5 hrs
Hence, 75% of the time the number of hours spent to prepare was less than 8.5hrs.
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Answer:
The probability of getting a sample with 80% satisfied customers or less is 0.0125.
Step-by-step explanation:
We are given that the results of 1000 simulations, each simulating a sample of 80 customers, assuming there are 90 percent satisfied customers.
Let = <u><em>sample proportion of satisfied customers</em></u>
The z-score probability distribution for the sample proportion is given by;
Z = ~ N(0,1)
where, p = population proportion of satisfied customers = 90%
n = sample of customers = 80
Now, the probability of getting a sample with 80% satisfied customers or less is given by = P( 80%)
P( 80%) = P(
) = P(Z
-2.24) = 1 - P(Z < 2.24)
= 1 - 0.9875 = <u>0.0125</u>
The above probability is calculated by looking at the value of x = 2.24 in the z table which has an area of 0.9875.
Using the normal distribution and the central limit theorem, it is found that there is a 0.0284 = 2.84% probability of finding a sample mean mass of 695g or below.
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Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
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- Mean of 700g means that
- Standard deviation of 21g means that
- Sample of 64, thus
- <u>For the sampling distribution of the sample mean</u>, the standard deviation is of
The probability of finding a sample mean mass of 695g or below is the p-value of Z when X = 695, thus:
By the Central Limit Theorem
has a p-value of 0.0284.
0.0284 = 2.84% probability of finding a sample mean mass of 695g or below.
A similar problem is given at brainly.com/question/22934264