Respuesta :
Answer:
a) 11.7% of students study for more than 10 hours per week.
b) 35.6% of student spends between 7 and 9 hours studying.
c) 1.6% of students spend fewer than 3 hours studying.
d) 5% of all A students spend studying 4.0455 or less hours during a week.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 7.5 hours
Standard Deviation, σ = 2.1 hours
We are given that the distribution of amount of time is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
a) P(students study for more than 10 hours per week)
P(x > 10)
[tex]P( x > 10) = P( z > \displaystyle\frac{10 - 7.5}{2.1}) = P(z > 1.1904)[/tex]
[tex]= 1 - P(z \leq 1.1904)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x > 10) = 1 - 0.883 = 0.117 = 11.7\%[/tex]
b) P(student spends between 7 and 9 hours studying.)
[tex]P(7 \leq x \leq 9) = P(\displaystyle\frac{7 - 7.5}{2.1} \leq z \leq \displaystyle\frac{9-7.5}{2.1}) = P(-0.2380 \leq z \leq 0.7142)\\\\= P(z \leq 0.7142) - P(z < -0.2380)\\= 0.762 - 0.406 = 0.356 = 35.6\%[/tex]
[tex]P(7 \leq x \leq 9) = 35.6\%[/tex]
c) P(students spend fewer than 3 hours studying)
[tex]P(x < 3) = P(z < \displaystyle\frac{3-7.5}{2.1}) = P(z < -2.1428)[/tex]
Calculating the value from the standard normal table we have,
[tex]P( x < 3) =0.016 = 1.6\%[/tex]
d) P(X < x) = 0.05
We have to find the value of x such that the probability is 0.035
[tex]P( X < x) = P( z < \displaystyle\frac{x - 7.5}{2.1})=0.05[/tex]
Calculation the value from standard normal z table, we have,
[tex]\displaystyle\frac{x - 7.5}{2.1} = -1.645\\x = 4.0455[/tex]
5% of all A students spend studying 4.0455 or less hours during a week.
You can use the standard normal distribution to get the z scores and with the help of it get the p values which gives the left sided probability of the z score.
The answers are:
- [tex]P( X > 10) = 0.117[/tex]
- [tex]P(7 < X < 9) = 0.0559[/tex]
- [tex]P(X < 3) = 0.0162[/tex]
- Below 4 hours, only 5% of all students of grade A spend studying.
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]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 context, the time spent in hours be tracked by a random variable X, then by the given data, we have:
[tex]X \sim N(7.5, 2.1)[/tex]
We've to evaluate
- [tex]P( X > 10) = 1 - P(X \leq 10)[/tex]
- [tex]P(7 < X < 9) = P(X \leq 9 ) - P( X \leq 7)[/tex]
- [tex]P(X < 3)[/tex]
- Only 5% students are below
Converting the distribution to standard normal variate, we get
[tex]Z = \dfrac{X - 7.5}{2.1}\\\\Z \sim N(0,1)[/tex]
Then the answers are obtained as
a) [tex]P( X > 10) = 1 - P(X \leq 10)[/tex]
[tex]P( X > 10) = 1 - P(X \leq 10) = 1 - P(Z < \dfrac{10 - 7.5}{2.1}) = 1 - P(Z < 1.19)[/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]
From z tables, the p value for z = 1.19 is 0.8830
Thus,
[tex]P( X > 10) = 1 - P(Z < 1.19) = 1 - 0.883 = 0.117[/tex]
b) [tex]P(7 < X < 9) = P(X \leq 9 ) - P( X \leq 7)[/tex]
[tex]P(7 < X < 9) = P(X \leq 9 ) - P( X \leq 7) =P(Z < 0.714) - P(Z < -0.24)[/tex]
We have p value for Z = 0.714 as 0.7611
and for Z = -0.24 as 0.4052
Thus,
[tex]P(7 < X < 9) =P(Z < 0.714) - P(Z < -0.24) = 0.7611 - 0.4052 = 0.0559[/tex]
c) [tex]P(X < 3)[/tex]
[tex]P(X < 3) = P(Z < \dfrac{3 - 7.5}{2.1}) = P(Z < -2.14)[/tex]
The p value for z = -2.14 is 0.0162
Thus,
[tex]P(X < 3) ) = P(Z < -2.14) = 0.0162[/tex]
d) Let the time be t hours, for which we have the only 5% students low than that.
That means P(X < t) = 0.05
[tex]P(X < t) = P(Z < \dfrac{t - 7.5}{2.1}) = 0.05[/tex]
Looking in z tables, we get at Z = -1.645, the p value is approx 0.05
Thus, we have:
[tex]\dfrac{t - 7.5}{2.1} = -1.645\\\\t = -1.645 \times 2.1 + 7.5 = 4.045 \approx 4[/tex]
The answers are:
- [tex]P( X > 10) = 0.117[/tex]
- [tex]P(7 < X < 9) = 0.0559[/tex]
- [tex]P(X < 3) = 0.0162[/tex]
- Below 4 hours, only 5% of all students of grade A spend studying.
Learn more about standard normal distribution here:
https://brainly.com/question/14684906
