1- Given standard normal distribution, find

a) P(Z <-2.33)
b) P(Z> 1.99)
C) P(-1.25 <Z<1.75)

2 Given standard normal distribution, find the z-score that

a) has 33% of the distribution's area to the right
b) corresponds to P5 (the 5th percentile);
c) has 75% of the distribution's area between -z and z.

3. The numbers of hours worked by medical residents is approximately normally distributed with a mean of 81.7 hours and a standard deviation of 6.9 hours.

a) What proportion of medical residents work less than 70 hours per week?

b) If 500 medical residents are selected, how many can be expected to work between 75 and 95 hours in a week?

c) What is the probability that a medical resident works more than 100 hours per week?

d) if the bottom 10% are excluded, what is the cut-off number of hours worked per week in order to be included?

e) what value of the number of hours worked per week represents the third quartile?

2. Originally Posted by hidaja16
1- Given standard normal distribution, find

a) P(Z <-2.33)

Mr F says: Pr(Z < -2.33) = Pr(Z > 2.33) = 1 - Pr(Z < 2.33). Now use your Standard Normal Distribution tables.

b) P(Z> 1.99)

Mr F says: Pr(Z > 1.99) = 1 - Pr(Z < 1.99). Now use your Standard Normal Distribution tables.

C) P(-1.25 <Z<1.75)

Mr F says: Pr(-1.25 < Z < 1.75) = Pr(Z < 1.75) - Pr(Z < -1.25). Get Pr(Z < -1.25) in a similar way to how part a) is done. Now use your Standard Normal Distribution tables.

2 Given standard normal distribution, find the z-score that

a) has 33% of the distribution's area to the right

Mr F says: Get the value of z* such that Pr(Z < z*) = 1 - 0.33 = 0.67 by using your Standard Normal Distribution tables in reverse.

b) corresponds to P5 (the 5th percentile);

Mr F says: Get the value of -z* such that Pr(Z < -z*) = 0.05. Note that Pr(Z < -z*) = 0.05 => Pr(Z > z*) = 0.05 => Pr(Z < z*) = 1 - 0.05 = 0.95. Get z* by using your Standard Normal Distribution tables in reverse.

c) has 75% of the distribution's area between -z and z.

Mr F says: Get the value of z* such that Pr(-z* < Z < z*) = 0.75. Note that Pr(-z* < Z < z*) = 0.75 => Pr(Z < -z*) + Pr(Z > z*) = 2Pr(Z > z*) = 0.25 => Pr(Z > z*) = 0.125 => Pr(Z < z*) = 1 - 0.125 = 0.875. Get z* by using your Standard Normal Distribution tables in reverse.

3. The numbers of hours worked by medical residents is approximately normally distributed with a mean of 81.7 hours and a standard deviation of 6.9 hours.

a) What proportion of medical residents work less than 70 hours per week?

Mr F says: Calculate Pr(X < 70) = Pr(Z < z*) where ${\color{red} z^* = \frac{70 - \mu}{\sigma} = \frac{70 - 81.7}{6.9}}$. Use Q1 to guide you here.

b) If 500 medical residents are selected, how many can be expected to work between 75 and 95 hours in a week?

Mr F says: Calculate Pr(75 < X < 95) = Pr(X < 95) - Pr(X < 75). use the help gven in a) and Q1 to guide you in this.

Then the expected value is ${\color{red} 500 \cdot \Pr(75 < X < 95)}$.

c) What is the probability that a medical resident works more than 100 hours per week?

Mr F says: Calculate Pr(X > 100) = 1 - Pr(X < 100) = 1 - Pr(Z < z*) where ${\color{red} z^* = \frac{100 - \mu}{\sigma} = \frac{100 - 81.7}{6.9}}$.

d) if the bottom 10% are excluded, what is the cut-off number of hours worked per week in order to be included?

Mr F says: You need the value of x* such that Pr(X < x*) = 0.1.

Get the value of -z* such that Pr(Z < -z*) = 0.1. Note that Pr(Z < -z*) = 0.1 => Pr(Z > z*) = 0.1 => Pr(Z < z*) = 1 - 0.1 = 0.9. Get z* by using your Standard Normal Distribution tables in reverse.

Then ${\color{red} -z^* = \frac{x^* - \mu}{\sigma} = \frac{x^* - 81.7}{6.9}}$. Now solve for x*.

e) what value of the number of hours worked per week represents the third quartile?

Mr F says: Find the value of x* such that Pr(X < x*) = 0.75. Use the above questions as a guide.
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