Question

Please help me solve this as many as you can. Thank you.

1/ Doing research for insurance rates, it is found that those aged 30 to 49 drive an average of 38.7 miles per day with a standard deviation of 6.7 miles. These distances are normally distributed. If a group of 60 drivers in that age group are randomly selected, what is the probability that the mean distance traveled each day is between 29.9 miles and 39.9 miles?

2/ A stock’s price fluctuations are approximately normally distributed with a mean of $29.51 and a standard deviation of $3.87. You decide to sell whenever the price reaches its highest 10% of values. What is the highest value you would still hold the stock?

Thank you again.

Answer 1

1) P(29.9 < $\bar x$ < 39.9)

= P((29.9 - $\mu$)/($\dpi{100} \sigma/\sqrt n$) < ($\bar x$ - $\mu$)/($\dpi{100} \sigma/\sqrt n$) < (39.9 - $\mu$)/($\dpi{100} \sigma/\sqrt n$))

= P((29.9 - 38.7)/(6.7/$\dpi{100} \sqrt 60$) < Z < (39.9 - 38.7)/(6.7/$\dpi{100} \sqrt 60$))

= P(-10.17 < Z < 1.39)

= P(Z < 1.39) - P(Z < -10)

= 0.9177 - 0

= 0.9177

2) P(X > x) = 0.10

or, P((X - $\dpi{100} \mu$)/$\dpi{100} \sigma$ > (x - $\dpi{100} \mu$)/$\dpi{100} \sigma$) = 0.10

or, P(Z > (x - 29.51)/3.87) = 0.10

or, P(Z < (x - 29.51)/3.87) = 0.90

or, (x - 29.51)/3.87 = 1.28

or, x = 1.28 * 3.87 + 29.51

or, x = 34.46