The probability that a data value is between 37 and 41 for this considered normal distribution is found being 0.6247.
How to get the z scores?
If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.
If we have
[tex]X \sim N(\mu, \sigma)[/tex]
(X is following normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex])
then it can be converted to standard normal distribution as
[tex]Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)[/tex]
(Know the fact that in continuous distribution, probability of a single point is 0, so we can write
[tex]P(Z \leq z) = P(Z < z) )[/tex]
Also, know that if we look for Z = z in z-tables, the p value we get is
[tex]P(Z \leq z) = \rm p \: value[/tex]
For this case, consider,
X = random variable which is normally distributed with mean of 38 and a standard deviation of 2.
Symbolically, we have:
[tex]X \sim N(\mu = 38,\sigma = 2)[/tex]
Then, the probability that a data value is between 37 and 41 is same as [tex]P(37 < X < 41)[/tex]
We can rewrite this as:
[tex]P(37 < X < 41) = P(X < 41) - P(X < 37)[/tex]
Converting X's distribution to standard normal distribution, we get:
[tex]P(X < 41) = P\left( Z < z = \dfrac{x - \mu}{\sigma} = \dfrac{41- 38}{2} \right) = P(Z < 1.5)[/tex]
Similarly, we get:
[tex]P(X < 37) = P\left( Z < z = \dfrac{x - \mu}{\sigma} = \dfrac{37 -38}{2} \right) = P(Z < -0.5)[/tex]
Thus, we have:
[tex]P(37 < X < 41) = P(X < 41) - P(X < 37) = P(Z < 1.5) - P(Z < -0.5)[/tex]
Using z-tables, we get the p-value for Z = 1.5 and Z = -0.5 as 0.9332, 0.3085 respectively.
Thus, we get:
[tex]P(37 < X < 41) =P(Z < 1.5) - P(Z < -0.5)\\P(37 < X < 41) = 0.9332 - 0.3085 = 0.6247[/tex]
Thus, the probability that a data value is between 37 and 41 for this considered normal distribution is found being 0.6247.
Learn more about z-score here:
https://brainly.com/question/21262765
Learn more about standard normal distribution here:
https://brainly.com/question/10984889
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