Quiz on "Modeling data distributions" (Unit 4, Jun 14th 2026)

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1:

Code:

mean = 170.4
sd = 10

l = 145
lz = (l - mean) / sd
print(lz)

import statistics
lz_area = statistics.NormalDist(mu=0, sigma=1).cdf(lz)
print(lz_area)

h = 171
hz = (h - mean) / sd
print(hz)

hz_area = statistics.NormalDist().cdf(hz)

area_req = round(hz_area - lz_area,4)
print(area_req)

---

2:

mean = 80
sd = 9

proportion = 0.4

import statistics
z = statistics.NormalDist().inv_cdf(proportion)

print(z)

x = z * sd + mean

print(x)

---

3:

Code:

mean = 13.1
sd = 1.5

sd1 = (mean - sd, mean + sd)
print(sd1)

sd2 = (mean - 2 * sd, mean + 2 * sd)
print(sd2)

sd3 = (mean - 3 * sd, mean + 3 * sd)
print(sd3)

sd2_area = 0.95
sd3_area = 0.997

area_req = (sd3_area - sd2_area) / 2

print(area_req)

percentage_wise = round(area_req * 100, 4)
print(percentage_wise)

out = """
(11.6, 14.6)
(10.1, 16.1)
(8.6, 17.6)
0.02350000000000002
2.35
"""

---

4:

Code:

b = 2
h = 0.6

area = 0.5 * b * h

percentage_of_area = area * 100

print(percentage_of_area)

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5:

Code:

mean_sales = 8000
sd_sales = 1500

mean_salary = 2000 + 0.3 * mean_sales

sd_salary = sd_sales * 0.3

print("mean_salary, sd_salary")
print(mean_salary, sd_salary)

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6:

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7:

area = 1
b = 6
h = area * 2 / b
print(h)

---

8:

def area_of_trapezium(b1, b2, h):
    return 0.5 * (b1 + b2) * h

b1 = 0.5
b2 = 0.75
h = 1

a = area_of_trapezium(b1, b2, h)
print(a)

print(round(a*100, 4))


b1 = 0.25
b2 = 0.5
h = 1

a = area_of_trapezium(b1, b2, h)
print(a)

print(round(a*100, 4))

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9:

---

10:

mean = 1497
sd = 322

proportion = 0.85

import statistics
z = statistics.NormalDist().inv_cdf(proportion)

x = z * sd + mean

print(round(x, 4))

See All: Questions For Statistics From Khan Academy
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