Respuesta :
Answer:
a) 68.2%
b) 31.8%
c) 2.3%
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 530
Standard Deviation, σ = 119
We are given that the distribution of math scores is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
a) P(test scores is between 411 and 649)
[tex]P(411 \leq x \leq 649)\\\\= P(\displaystyle\frac{411 - 530}{119} \leq z \leq \displaystyle\frac{649-530}{119})\\\\= P(-1 \leq z \leq 1)\\\\= P(z \leq 1) - P(z < -1)\\= 0.841 - 0.159 = 0.682 = 68.2\%[/tex]
b) P(scores is less than 411 or greater than 649)
[tex]P(x < 411 < x, x > 649)\\=1 = P(411 \leq x \leq 649)\\=1 - 0.682\\=0.318 = 31.8\%[/tex]
c) P(score greater than 768)
P(x > 768)
[tex]P( x > 768) = P( z > \displaystyle\frac{768 - 530}{119}) = P(z > 2)[/tex]
[tex]= 1 - P(z \leq 2)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x > 768) = 1 -0.977 = 0.023 = 2.3\%[/tex]
You can convert the given normal distribution to standard normal distribution and then use the z tables to find the needed probabilities.
Rounding to one places of decimal, we get the answers as:
- [tex]P( 411 < X < 649 ) \approx 0.6826 = 68.2\%[/tex]
- [tex]P(X < 411) + P(X > 649) \approx 0.3174 \approx 31.7\%[/tex]
- [tex]P(X > 768) \approx 0.0228 \approx 2.3\%[/tex]
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]
Using the z scores will help to find the probabilities from the z tables(available online).
Let for the given test, the test scores be tracked by a random variable X, then by the given data, we have:
[tex]X \sim N(530, 119)[/tex]
The needed probabilities are
- [tex]P( 411 < X < 649 ) = P(X < 649) - p(X < 411)\\[/tex]
- [tex]P(X < 411) + P(X > 649) = 1 - P(411 \leq X \leq 649) = 1 - P(411 < X < 649)[/tex]
- [tex]P(X > 768) = 1 - P(X \leq 768)[/tex]
Converting the distribution to standard normal variate, the probabilities become
[tex]Z = \dfrac{X - 530}{119}\\\\Z \sim N(0, 1)[/tex]
The probabilities convert to
a) [tex]P( 411 < X < 649 ) = P(X < 649) - p(X < 411)\\[/tex]
[tex]P(\dfrac{411 - 530}{119} < Z < \dfrac{ 649 - 530}{119}) = P(-1 < Z < 1) = P(Z < 1) - P(Z < -1)[/tex]
(Know the fact that in continuous distribution, probability of a single point is 0, so we can write [tex]P(Z < a) = P(Z \leq a)[/tex] )
Also, know that if we look for Z = a in z tables, the p value we get is [tex]P(Z \leq a) = p \: value[/tex]
The p value at Z = 1 is 0.8413 and at Z = -1 is 0.1587,
Thus, [tex]P(411 < X < 649) = P(Z < 1) - P(Z < -1) = 0.8413 - 0.1587 = 0.6826[/tex]
b) [tex]P(X < 411) + P(X > 649) = 1 - P(411 \leq X \leq 649) = 1 - P(411 < X < 649)\\\\P(X < 411) + P(X > 649) = 1 - 0.6826 = 0.3174[/tex]
c) [tex]P(X > 768) = 1 - P(X \leq 768)[/tex]
[tex]P(X > 768) = 1 - P(Z < \dfrac{768 - 530}{119}) = 1 - P(Z < 2) = 1 - 0.9772 = 0.0228[/tex]
Rounding to one places of decimal, we get the answers as:
- [tex]P( 411 < X < 649 ) \approx 0.6826 = 68.2\%[/tex]
- [tex]P(X < 411) + P(X > 649) \approx 0.3174 \approx 31.7\%[/tex]
- [tex]P(X > 768) \approx 0.0228 \approx 2.3\%[/tex]
Learn more about standard normal distribution here:
https://brainly.com/question/21262765
