Respuesta :
Answer:
(a) Probability that a visually impaired student gets less than 6.5 hours of sleep is 0.1390.
(b) P(6.9 hours < X < 10.6 hours) = 0.6172
(c) Forty percent of students get less than 8.26 hours of sleep on a typical day.
Step-by-step explanation:
We are given that Researchers found that visually impaired students averaged 8.8 hours of sleep, with a standard deviation of 2.12 hours.
Assume that the number of hours of sleep for these visually impaired students is normally distributed.
Let X = number of hours of sleep for these visually impaired students
So, X ~ Normal([tex]\mu=8.8,\sigma^{2} = 2.12^{2}[/tex])
The z-score probability distribution for normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = mean hours of sleep = 8.8 hours
[tex]\sigma[/tex] = standard deviation = 2.12 hours
(a) Probability that a visually impaired student gets less than 6.5 hours of sleep is given by = P(X < 6.5 hours)
P(X < 6.5 hours) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{6.5-8.8}{2.12}[/tex] ) = P(Z < -1.085) = 1 - P(Z [tex]\leq[/tex] 1.085)
= 1 - 0.8610 = 0.1390
The above probability is calculated by looking at the value of x = 1.085 in the z table which will lie in between x = 1.08 and x = 1.09.
(b) Probability that a visually impaired student gets between 6.9 and 10.6 hours of sleep is given by = P(6.9 hours < X < 10.6 hours)
P(6.9 hours < X < 10.6 hours) = P(X < 10.6 hours) - P(X [tex]\leq[/tex] 6.9 hours)
P(X < 10.6 hours) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{10.6-8.8}{2.12}[/tex] ) = P(Z < 0.85) = 0.80234
P(X [tex]\leq[/tex] 6.9 hours) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{6.9-8.8}{2.12}[/tex] ) = P(Z [tex]\leq[/tex] -0.896) = 1 - P(Z < 0.896)
= 1 - 0.81487 = 0.18513
The above probability is calculated by looking at the value of x = 0.85 and x = 0.896 in the z table which will lie in between x = 0.89 and x = 0.90.
Therefore, P(6.9 hours < X < 10.6 hours) = 0.80234 - 0.18513 = 0.6172
(c) We have to find that Forty percent of students get less than how many hours of sleep on a typical day, that is;
P(X < [tex]x[/tex]) = 0.40 {where [tex]x[/tex] is the hours of sleep}
P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{x-8.8}{2.12}[/tex] ) = 0.40
P(Z < [tex]\frac{x-8.8}{2.12}[/tex] ) = 0.40
Now, in the z table the critical value of x for which the probability area is less than 40% is -0.2533, that means;
[tex]\frac{x-8.8}{2.12}[/tex] = -0.2533
[tex]x-8.8 = -0.2533 \times 2.12[/tex]
[tex]x[/tex] = 8.8 - 0.536996 = 8.26
Therefore, Forty percent of students get less than 8.26 hours of sleep on a typical day.
