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Sagot :
Answer:
The interval is [98,132]
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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.
Normal with mean 115 and standard deviation 25.
This means that [tex]\mu = 115, \sigma = 25[/tex]
Determine the interval of values that is centered at the mean and for which 50% of the students have IQ's in that interval.
Between the 50 - (50/2) = 25th percentile and the 50 + (50/2) = 75th percentile.
25th percentile:
X when Z has a p-value of 0.25, so X when Z = -0.675.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.675 = \frac{X - 115}{25}[/tex]
[tex]X - 115 = -0.675*25[/tex]
[tex]X = 98[/tex]
75th percentile:
X when Z has a p-value of 0.75, so X when Z = 0.675.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.675 = \frac{X - 115}{25}[/tex]
[tex]X - 115 = 0.675*25[/tex]
[tex]X = 132[/tex]
The interval is [98,132]
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