Solution;
From the given question, we have
[tex]\begin{gathered} \mu=7 \\ \sigma=1 \end{gathered}[/tex]
To evaluate the proportion of people that spend at least 5 hours per day on their home computers,
Step 1: Evaluate the z score value.
The z score value is expressed as
[tex]\begin{gathered} z=\frac{x-\mu}{\sigma} \\ where \\ x\implies sample\text{ value} \\ \mu\implies mean\text{ value} \\ \sigma\implies standard\text{ deviation} \end{gathered}[/tex]
In this case,
[tex]x=5[/tex]
thus, we have the z score value to be evaluated as
[tex]z=\frac{5-7}{1}=-2[/tex]
Step 2: Evaluate the probability that people will spend at least 5 hours per day on their home computers,
From the normal distribution table, we have
This implies that
[tex]P(x\ge5)=P(z\ge-2)=0.977249868052[/tex]
Hence, the proportion that spends at least 5 hours per day on their home computers is evaluated to be
[tex]\begin{gathered} 0.977249868052\times100 \\ =97.7249868052 \\ \approx97.72\%\text{ \lparen nearest hundredth\rparen} \end{gathered}[/tex]
