Let X = a score on the final exam. X ~ N(63, 5), where μ = 63 and σ = 5

Draw a graph.

Then, find P(x > 65).

P(x > 65) = 0.3446

The probability that any student selected at random scores more than 65 is 0.3446.

Go into 2nd DISTR.

After pressing 2nd DISTR, press2:normalcdf.

The syntax for the instructions are as follows: normalcdf(lower value, upper value, mean, standard deviation) For this problem: normalcdf(65,1E99,63,5) = 0.3446. You get 1E99 (= 10⁹⁹) by pressing 1, the EE key (a 2nd key) and then 99. Or, you can enter10^99 instead. The number 10⁹⁹ is way out in the right tail of the normal curve. We are calculating the area between 65 and 10⁹⁹. In some instances, the lower number of the area might be –1E99 (= –10⁹⁹). The number –10⁹⁹ is way out in the left tail of the normal curve.

[latex]\displaystyle{z}=\frac{{{65}-{63}}}{{5}}={0.4}[/latex]

Area to the left is 0.6554.
P(x > 65) = P(z > 0.4) = 1 – 0.6554 = 0.3446 

Draw a graph.

Then find P(x < 85), and shade the graph.

Using a computer or calculator, find P(x < 85) = 1.

normalcdf(0,85,63,5) = 1 (rounds to one).

The probability that one student scores less than 85 is approximately one (or 100%).

Find the 90th percentile. For each problem or part of a problem, draw a new graph. Draw the x-axis. Shade the area that corresponds to the 90th percentile.

Let k = the 90th percentile. The variable k is located on the x-axis. P(x < k) is the area to the left of k. The 90th percentile k separates the exam scores into those that are the same or lower than k and those that are the same or higher. Ninety percent of the test scores are the same or lower than k, and ten percent are the same or higher. The variable k is often called a critical value.

k = 69.4

The 90th percentile is 69.4. This means that 90% of the test scores fall at or below 69.4 and 10% fall at or above. To get this answer on the calculator, follow this step:

Find the 70th percentile.

Draw a new graph and label it appropriately. k = 65.6

The 70th percentile is 65.6. This means that 70% of the test scores fall at or below 65.5 and 30% fall at or above.

invNorm(0.70,63,5) = 65.6

a. Let X = the amount of time (in hours) a household personal computer is used for entertainment. X ~ N(2, 0.5) where μ = 2 and σ = 0.5.

Find P(1.8 < x < 2.75).

The probability for which you are looking is the area between x = 1.8 and x = 2.75. P(1.8 < x < 2.75) = 0.5886

normalcdf(1.8,2.75,2,0.5) = 0.5886

The probability that a household personal computer is used between 1.8 and 2.75 hours per day for entertainment is 0.5886.

b. To find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment, find the 25th percentile, k, where P(x < k) = 0.25.

invNorm(0.25,2,0.5) = 1.66

The maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment is 1.66 hours.

normalcdf(23,64.7,36.9,13.9) = 0.8186

normalcdf(–1099,50.8,36.9,13.9) = 0.8413

• invNorm(0.80,36.9,13.9) = 48.6
• The 80th percentile is 48.6 years.
• 80% of the smartphone users in the age range 13 – 55+ are 48.6 years old or less.

a.

• IQR = Q₃ – Q₁Calculate
• Q₃ = 75th percentile and Q₁ = 25th percentile.
• invNorm(0.75,36.9,13.9) = Q₃ = 46.2754
• invNorm(0.25,36.9,13.9) = Q₁ = 27.5246
• IQR = Q₃ – Q₁ = 18.7508

b.

• Find k where P(x > k) = 0.40 (“At least” translates to “greater than or equal to.”)
• 0.40 = the area to the right.
• Area to the left = 1 – 0.40 = 0.60.
• The area to the left of k = 0.60.
• invNorm(0.60,36.9,13.9) = 40.4215.
• k = 40.42.
• Forty percent of the ages that range from 13 to 55+ are at least 40.4 years.

normalcdf(6,10^99,5.85,0.24) = 0.2660

b.

• 1 – 0.20 = 0.80
• The tails of the graph of the normal distribution each have an area of 0.40.
• Find k1, the 40th percentile, and k2, the 60th percentile (0.40 + 0.20 = 0.60).
• k1 = invNorm(0.40,5.85,0.24) = 5.79 cm
• k2 = invNorm(0.60,5.85,0.24) = 5.91 cm

c. 6.16: Ninety percent of the diameter of the mandarin oranges is at most 6.16 cm.