Question

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 be the mean number of hours spent on the computer for this group.
Find the probability that y is between 6.2 and 6.9 hours.

Answer 1

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 = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$  ~ N(0,1)

where, $\mu$ = population mean hours spent = 6 hours

            $\sigma$ = population standard deviation = 2 hours

            n = sample of students = 80

Here $\bar X$ = 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 $\leq$ 6.2 hours)

 P(y < 6.9 hours) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{6.9-6}{\frac{2}{\sqrt{80} } }$ ) = P(Z < 4.02) = 0.99997

 P(y $\leq$ 6.2 hours) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{6.2-6}{\frac{2}{\sqrt{80} } }$ ) = P(Z $\leq$ 0.89) = 0.81327                          

{Now, in the z table the P(Z $\leq$ 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.