Question 4
Topic: SL 4.2
Given
The diameters of the tubes, Dmm, are normally distributed with mean 32 and standard deviation s. The interquartile range of the diameters is 0.28.
Find
the value of \(\sigma \).
▶️Answer/Explanation
Answer:
METHOD 1
\(Q_1\) = 31.86 OR \(Q_3\) = 32.14 recognition that the area under the normal curve below \(Q_1\) or above \(Q_3\) is 0.25 OR the area between \(Q_1\) and \(Q_3\) is 0.5 (seen anywhere including on a diagram)
EITHER
equating an appropriate correct normal CDF function to its correct probability (0.25 or 0.5 or 0.75)
OR
z = −0.674489… OR z = 0.674489… (seen anywhere)
-0.674489… = \(\frac{31.86 – 32}{\sigma }\) OR 0.674489… = \(\frac{32.14 – 32}{\sigma }\)
THEN
0.207564…
\(\sigma = 0.208\) (mm)
METHOD 2
recognition that the area under the normal curve below \(Q_1\) or above \(Q_3\) is 0.25 OR the area between \(Q_1\) and \(Q_3\) is 0.5 (seen anywhere including on a diagram)
z = −0.674489… OR z = 0.674489…
\((Q_1=) 32 – 0.674489… \sigma \) OR \((Q_3 =) 32 + 0.674489… \sigma \)
\((Q_3 – Q_1 =) 2 \times 0.674489… \sigma \)
2 \(\times \) 0.674489… \(\sigma \) = 0.28
0.207564…
\(\sigma = 0.208\) (mm)
The interquartile range (IQR) is a measure of statistical dispersion, which is the difference between the third quartile (Q3) and the first quartile (Q1) of a dataset. For a normal distribution, the IQR is related to the standard deviation \((\sigma)\) by a factor of approximately 1.35.
\(\text{IQR} = Q_3 – Q_1 = 1.349\sigma\)
Where:
- IQR is the interquartile range
- Q3 is the 75th percentile
- Q1 is the 25th percentile
- σ is the standard deviation
The interquartile range of the diameters is 0.28.
We can substitute the known IQR value into the equation: \(0.28 = 1.349\sigma \)
\(\sigma = \frac{0.28}{1.349} \)
\(\sigma \approx 0.208\)
Question 5
Topic : SL 1.9
Given
The coefficient of \(x^6\) in the expansion of \((ax^3 + b)^8\) is 448.
The coefficient of \(x^6\) in the expansion of (\(ax^3 + b)^10\) is 2880.
Find
value of a and the value of b, where a, b > 0.
▶️Answer/Explanation
Answer:
product of a binomial coefficient, a power of \(ax^3\) and a power of b seen
evidence of correct term chosen
for n = 8 : r = 2 (or r = 6) OR for n =10 : r = 2 (or r = 8)
correct equations (may include powers of x)
attempt to solve their system in a and b algebraically or graphically
b = 2 ; a = \(\frac{1}{2}\)
Step 1: Express the coefficients in terms of a and b using the binomial expansion formula.
\(\text{For } (ax^3 + b)^8, \text{ coefficient of } x^6 = (C_{8}^{2}\textrm{}) \times a^6 \times b^2 = 448\)
\(\text{For } (ax^3 + b)^{10}, \text{ coefficient of } x^6 = (C_{10}^{4}) \times a^6 \times b^4 = 2880\)
Step 2: Substitute the given values of the coefficients & Simplify the combinatorial expressions.
\(448 = 28 \times a^6 \times b^2 \)
\(2880 = 210 \times a^6 \times b^4 \)
\(\frac{2880}{448} = \frac{210 \times b^4}{28 \times b^2}\)
\(7.5 = \frac{210}{28} \times b^2 \)
\(b^2 = \frac{7.5 \times 28}{210} = 1\)
b = 1
\(448 = 28 \times a^6 \times 1^2 \)
\(a^6 = \frac{448}{28} = 16 \)
a = 2
Therefore, the value of a is 2, and the value of b is 1.