IB Mathematics SL 4.9 normal distribution AA SL Paper 2- Exam Style Questions- New Syllabus
A bakery makes two types of muffins: chocolate muffins and banana muffins.
The weights, \( C \) grams, of the chocolate muffins are normally distributed with a mean of 62g and standard deviation of 2.9g.
a) Find the probability that a randomly selected chocolate muffin weighs less than 61g [3]
b) In a random selection of 12 chocolate muffins, find the probability that exactly 5 weigh less than 61g [3]
The weights, \( B \) grams, of the banana muffins are normally distributed with a mean of 68g and standard deviation of 3.4g. Each day 60% of the muffins made are chocolate.
On a particular day, a muffin is randomly selected from all those made at the bakery.
ci) Find the probability that the randomly selected muffin weighs less than 61g [3]
cii) Given that a randomly selected muffin weighs less than 61g, find the probability that it is chocolate [3]
d) The machine that makes the chocolate muffins is adjusted so that the mean weight remains the same but their standard deviation changes to \( \sigma \) g. The probability that the weight of a randomly selected muffin is less than 61g is now 0.157. Find the value of \( \sigma \) [4]
▶️ Answer/Explanation
a) Use normal distribution:
\( C \sim N(62, 2.9^2) \)
\( z = \frac{61 – 62}{2.9} = \frac{-1}{2.9} \approx -0.3448 \)
\( P(Z < -0.3448) \approx 1 – P(Z < 0.3448) \)
\( P(Z < 0.3448) \approx 0.6352 \)
\( P(C < 61) \approx 1 – 0.6352 = 0.3648 \)
Rounded to three decimal places:
\( \approx 0.365 \) [3]
b) Use binomial distribution:
\( X \sim B(12, 0.365) \)
\( P(X = 5) = \binom{12}{5} \times 0.365^5 \times 0.635^7 \)
\( \binom{12}{5} = 792 \)
\( 0.365^5 \approx 0.00064757 \)
\( 0.635^7 \approx 0.041595 \)
\( P(X = 5) \approx 792 \times 0.00064757 \times 0.041595 \approx 0.21351 \)
Rounded to three decimal places:
\( \approx 0.214 \) [3]
ci) Use total probability:
\( B \sim N(68, 3.4^2) \)
\( z = \frac{61 – 68}{3.4} = \frac{-7}{3.4} \approx -2.0588 \)
\( P(Z < -2.0588) \approx 0.01977 \)
\( P(\text{Chocolate}) = 0.6 \), \( P(\text{Banana}) = 0.4 \)
\( P(C < 61) \approx 0.365 \)
\( P(B < 61) \approx 0.01977 \)
\( P(\text{<61g}) = 0.6 \times 0.365 + 0.4 \times 0.01977 \)
\( \approx 0.219 + 0.007908 \approx 0.226908 \)
Rounded to three decimal places:
\( \approx 0.227 \) [3]
cii) Use conditional probability:
\( P(\text{Chocolate} | \text{<61g}) = \frac{P(\text{Chocolate} \cap \text{<61g})}{P(\text{<61g})} \)
\( P(\text{Chocolate} \cap \text{<61g}) = 0.6 \times 0.365 \approx 0.219 \)
\( P(\text{<61g}) \approx 0.226908 \)
\( P(\text{Chocolate} | \text{<61g}) \approx \frac{0.219}{0.226908} \approx 0.96525 \)
Rounded to three decimal places:
\( \approx 0.965 \) [3]
d) Use total probability with new standard deviation:
\( C \sim N(62, \sigma^2) \)
\( P(\text{<61g}) = 0.6 \times P(C < 61) + 0.4 \times 0.01977 = 0.157 \)
\( P(B < 61) \approx 0.01977 \)
\( 0.6 \times P(C < 61) + 0.007908 \approx 0.157 \)
\( 0.6 \times P(C < 61) \approx 0.149092 \)
\( P(C < 61) \approx \frac{0.149092}{0.6} \approx 0.248487 \)
\( z = \frac{61 – 62}{\sigma} = \frac{-1}{\sigma} \)
\( P\left(Z < \frac{-1}{\sigma}\right) = 0.248487 \)
\( P(Z < -z) = 0.751513 \)
\( -z \approx -0.6792 \)
\( z \approx 0.6792 \)
\( \frac{-1}{\sigma} = -0.6792 \)
\( \sigma \approx \frac{1}{0.6792} \approx 1.47214 \)
Rounded to two decimal places:
\( \sigma \approx 1.47 \) g [4]