Detailed Answer

a) 68% of the data is within \( \sigma \) of the mean (between \( 4,750 \) and \( 5,250 \)), and 95% within \( 2\sigma \) (between \( 4,500 \) and \( 5,500 \)).

b) The variance is \( 225 \), so the standard deviation \( \sigma \) is \( 25 \). 68% falls within \( \sigma \) of the mean (between \( 350 \) and \( 400 \)), and 99.7% within \( 3\sigma \) (between \( 300 \) and \( 450 \)).

Detailed Answer

a)

\[ P(x \geq 11) = P(X=11) + P(X=12) + P(X=13) + P(X=14)= \]

\[ = 0.12 + 0.08 + 0.06 + 0.02 = 0.28 \]

b) \( P(x < 8) = P(x=6) + P(x=7) = 0.04 + 0.07 = 0.11 \)

c) \( P(8 \leq x \leq 10) = 1 - P(x \geq 11) - P(x < 8) = 1 - 0.28 - 0.11 = 0.61 \)

or: \( P(8 \leq x \leq 10) = P(x=8) + P(x=9) + P(x=10) = 0.20 + 0.22 + 0.19 = 0.61 \)

d) \( x \) can only take values up to 14 according to the probability distribution. It’s not possible to get any higher value, so \( P(x > 14) = 0 \).

Detailed Answer

a) This can be taken directly from the GDC:

\[ P(X<700) = 0.0228 \]

b) Here, we need to recognize that this will be a binomial distribution with the probability we found in part (a), 20 possibilites, and exactly 1 success:

\[ P(X=1) = 0.294 \]

Detailed Answer

a) The standard deviation is $1200, so around 68% of people are between $2,800 and $5,200, as these are \( \mu - \sigma \) and \( \mu + \sigma \). This can also be obtained on the calculator, where we find 68.27%.

This is the same as a binomial distribution with \( p = 0.6827 \) and \( n = 2 \). We want to have two successes out of two: \( P(X=2) = 0.6827 \times 0.6827 \approx 0.466 \)

b) We must find the inverse normal. With a calculator, we find that you have to be above $5540.

c) We find the inverse normal on a calculator, so: \( d = 4810 \, \text{USD} \)

d) With a calculator, we find \( P(x \leq 3000) = 0.202 \)

Detailed Answer

a)

b) We want to find the weight with only 25% of data before it. We use a calculator to find the inverse normal. We find \( t = 46.5 \, \text{g} \).

c) This is the weight with 75% of the data behind it. With the calculator we find \( t = 51.1 \, \text{g} \).

d) We find with the calculator: \( P(x \leq 46) = 0.2051 \)

\(50 \times 0.2051 = 10.255 \), so we expect 10 eggs meet that criterion.

e) With the calculator: \( P(45 \leq x \leq 55) = 0.834 \)

\(50 \times 0.834 = 41.7 \), so we expect 42 eggs.

Detailed Answer

a) (i) & (ii) By plugging the values into the calculator we get:

\[P(X < 10) = 0.0488 \]

\[P(X > 15) = 0.00621 \]

b) With the inverse normal function in the calculator we get:

\[ d= 13\]

c) By plugging the values into the calculator we get:

\[P(X < 8.5) = 0.00177 \]

Detailed Answer

a)

\[P(40000 < X < 48000) = 0.0103 \]

b) Using the inverse normal function in the calculator with the area of 0.2 we can find that the upper and lower limits are:

\[ lower = 23353.24384\]

\[ upper = 26646.75616\]

From that we can find the value for \( q \):

\[ q = 25000 - 23353.24384 \approx 1646.76\]

Detailed Answer

a) \(^{10}C_{10} * 0.98^{10} = 0.817\)

b) \(^{10}C_{8} * 0.98^8 * 0.02^2 + ^{10}C_{9} * 0.98^9 * 0.02^1 + ^{10}C_{10} * 0.98^{10} = 0.991\)

c) \(0.3 * 0.98 = 0.294\)

d) \(P(X ≥ 3) = 1 – P(X ≤ 2) = 1 – (P(X=1) + P(X=2) = \)

\(1-(^{10}C_{2} * 0.294^2 * (1 - 0.294)^8 + ^{10}C_{1} * 0.294^1 * (1 - 0.294)^9) = 0.601\)

Detailed Answer

a)

\[ P(X=210) = 0.0147 \]

b)

\[ P(X>240) = 0.212 \]

c) It will be a binomial distribution with the probability obtained in part (b):

\[ P(X<4) = 0.910 \]

d) We have to first calculate the probability of a cheeseburger to weigh more than 240g:

\[ P(X>240) = 0.356 \]

There will be two possibilites of a burger to weigh more than 240g, it can either be a cheeseburger or a hamburger, so the probabilities have to be added:

\[ 0.55 \times 0.356 + 0.45 \times 0.212 = 0.291\]

e)

\[ P(C|>240) = \frac{P(C \ \cap >240)}{P(>240)} = \frac{0.55 \times 0.356}{0.291} = 0.672 \]

Detailed Answer

a) Solving \( 5a = 1 \), we get:

\[ a = 0.2 \]

b) Finding \( E(X) \):

\[ E(X) = 1a + 2(2a) + 3a + 4a = 12a \]

Substituting \( a = 0.2 \):

\[ E(X) = 12(0.2) = 6 \]

c) i) The range of \( c \) is:

\[ 0 \leq c \leq 1 \]

ii) For \( c = 0 \), \( b = \frac{1}{4} \), and for \( c = 1 \), \( b = 0 \), so:

\[ 0 \leq b \leq \frac{1}{4} \]

d) Finding the range of \( E(Y) \):

\[ E(Y) = 1b + 2b + 3(2b) + 4c \]

For \( c = 0, b = \frac{1}{4} \):

\[ E(Y) = 1(0.25) + 2(0.25) + 3(0.25) + 4(0) = 1.75 \]

For \( c = 1, b = 0 \):

\[ E(Y) = 1(0) + 2(0) + 3(0) + 4(1) = 4 \]

Thus:

\[ 1.75 \leq E(Y) \leq 4 \]

e) Probability that Barbara wins:

Winning combinations: \( (1,2), (1,3), (1,4), (2,3), (2,4), (3,4) \)

\[ P(A < B) = ab + 2ab + ac + 4ab + 2ac + ac = 7ab + 4ac \]

Since \( 4b + c = 1 \), we substitute \( c = 1 - 4b \):

\[ 7ab + 4a(1 - 4b) = 0.5 \]

Substituting \( a = 0.2 \) and solving:

\[ b = \frac{1}{6}, \quad c = \frac{1}{3} \]

Final \( E(Y) \):

\[ E(Y) = 1 \left( \frac{1}{6} \right) + 2 \left( \frac{1}{6} \right) + 3 \left( \frac{1}{6} \right) + 4 \left( \frac{1}{3} \right) \]

\[ E(Y) = \frac{1}{6} + \frac{2}{6} + \frac{3}{6} + \frac{4}{3} = \frac{7}{3} \]

Detailed Answer

a) Let \( X \) be the mass of packs of flour.

\[ P(X < 498) \approx 0.159 \]

b) We use the probability we found in part (a):

\[ 0.159 \times 100 = 15.9 \approx 16 \]

c) Using the conditional probability formula:

\[ P(X > 502 \mid X \geq 498) = \frac{P(X > 502)}{P(X \geq 498)} \]

\[ P(X > 502 \mid X \geq 498) = \frac{0.158655\ldots}{1 - 0.158655\ldots} \approx 0.189 \]

Detailed Answer

a)

\[ 1 - (0.23 + (h - 0.27) + 0.64 + (0.28 - 2h^2)) = 2h^2 \]

\[ 2h^2 - h + 0.12 = 0 \]

b) Using the GDC:

\[ h = 0.2 \text{ and } h = 0.3 \]

Since probability cannot be negative:

\[ h - 0.27 < 0 \text{ when } h = 0.2 \]

Thus, we select: \( h = 0.3 \).

c) Substituting \( h = 0.3 \) into the expected value equation:

\[ E(X) = 1(0.23) + 2(0.3 - 0.27) + 3(0.64) + 4(0.28 - 2(0.3)^2) \]

\[ E(X) = 2.61 \]