Respuesta :
Answer:
Probability that y is between 6.2 and 6.9 hours is 0.1867.
Step-by-step explanation:
We are given that the number of hours per week that high school seniors spend on computers is normally distributed, with a mean of 6 hours and a standard deviation of 2 hours.
80 students are chosen at random.
Let y = sample mean number of hours spent on the computer for this group.
The z-score probability distribution for sample mean is given by;
Z = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean hours spent = 6 hours
[tex]\sigma[/tex] = population standard deviation = 2 hours
n = sample of students = 80
Here [tex]\bar X[/tex] = y
The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.
So, probability that y is between 6.2 and 6.9 hours is given by = P(6.2 hours < y < 6.9 hours) = P(y < 6.9 hours) - P(y [tex]\leq[/tex] 6.2 hours)
P(y < 6.9 hours) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{6.9-6}{\frac{2}{\sqrt{80} } }[/tex] ) = P(Z < 4.02) = 0.99997
P(y [tex]\leq[/tex] 6.2 hours) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] [tex]\leq[/tex] [tex]\frac{6.2-6}{\frac{2}{\sqrt{80} } }[/tex] ) = P(Z [tex]\leq[/tex] 0.89) = 0.81327
{Now, in the z table the P(Z [tex]\leq[/tex] x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 4.02 and x = 0.89 in the z table which has an area of 0.99997 and 0.81327 respectively.}
Therefore, P(6.2 hours < y < 6.9 hours) = 0.99997 - 0.81327 = 0.1867
Hence, the probability that y is between 6.2 and 6.9 hours is 0.1867.
