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2 answers:
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We have
Mean, μ = 25 minutes
Standard Deviation, σ = 6.1
X = 30 minutes
The probability we are looking for is shown in the first diagram; it's the area on the right of X=30
We need to standardized the value X=30 using the formula
rounded to two dp
The z-table is shown on the second diagram only gives the probability when P(Z<z), so to work out the probability when P(Z>z), we do 1-P(Z<z)
P(Z>0.82) = 1 - P(Z<z) = 1 - 0.7939 = 0.2061
Mean, μ = 25 minutes
Standard Deviation, σ = 6.1
X = 30 minutes
The probability we are looking for is shown in the first diagram; it's the area on the right of X=30
We need to standardized the value X=30 using the formula
The z-table is shown on the second diagram only gives the probability when P(Z<z), so to work out the probability when P(Z>z), we do 1-P(Z<z)
P(Z>0.82) = 1 - P(Z<z) = 1 - 0.7939 = 0.2061
Answer:
a) The mean is 18 and the standard deviation is 1.79.
b) The interpretation is that the standard deviation of the sample means of groups of 20 students will be of 1.79, which is the sample error, which is different from the population standard deviation.
c) 0.2005 = 20.05% probability that a sample of 20 students has a mean score of 19.5 or more.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
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 estabilishes 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.
Mean of 18 and a standard deviation of 8.
This means that
Sample of 20:
This means that
(a) Calculate the mean and standard deviation of the sampling distribution of x¯.
By the Central Limit Theorem, the mean is 18 and the standard deviation is 1.79.
(b) Interpret the standard deviation from part (a).
The interpretation is that the standard deviation of the sample means of groups of 20 students will be of 1.79, which is the sample error, which is different from the population standard deviation.
(c) Find the probability that a sample of 20 students has a mean score of 19.5 or more.
This is 1 subtracted by the pvalue of Z when X = 19.5. So
By the Central Limit Theorem
has a pvalue of 0.7995
1 - 0.7995 = 0.2005
0.2005 = 20.05% probability that a sample of 20 students has a mean score of 19.5 or more.