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1 answer:
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Answer:
This measure is just 0.17 standard deviations from the mean, so we should not be surprised.
Step-by-step explanation:
Normal probability distribution
When the distribution is normal, we use 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.
If the z-score is lower than -2, or higher than 2.5, the score of X is considered unusual.
Subtraction of normal variables:
When we subtract normal variables, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.
Let xS – xL represent the sampling distribution.
Mean s 6, means L 4. So
Standard deviation s is 2.5, for L is 1.5. So
Should we be surprised if the sample mean housebroken age for the small breed dogs is at least 2.5 months more than the sample mean housebroken age for the large breed dogs? Explain your answer.
We have to find the z-score for X = 2.5. So
This measure is just 0.17 standard deviations from the mean, so we should not be surprised.
Probability of an adult who needs correction = 82% or 0.82
So, probability of an adult who does not needs correction = 1 - 0.82 = 0.18
If 12 adults are randomly selected, the probability that at least 11 of them need correction for their eyesight is:
12C11 + 12C12
= 12*0.11*0.18 + 1*0.09*1
= 0.238+0.09
= 0.328 or 0.33
And, yes, 11 is significantly a high number of adults requiring eyesight correction.