Detailed Answer

b) and c) can be counted easily. The others have an infinite number of possible values.

Detailed Answer

a) \( 0.26 + 0.37 + 0.31 + 0.26 = 1.2 \), but the sum of the probabilities should be \( 1 \).

b) \( 0.4 + 0.3 + 0.2 + 0.05 + 0.05 = 1 \)

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)

\[ 0.5 \times E(X) = 0.5 \times n \times p \]

\[ 0.5 \times 12 \times 0.6 = 0.5 \times 7.2 = 3.6 \]

b) \(\frac{3.5}{0.5} = 7\)

\[P(x=7) = \binom{12}{7} \times 0.67 \times 0.45 = 0.227 \]Also given by the calculator.

c) \(\frac{4}{0.5} = 8\)

\[P(x>9) = P(x=10) + P(x=11) + P(x=12) = 0.0833 \]Also given by the calculator.

d) \(\frac{12}{0.5} = 24\)

We want to have 24 views, so \( E(X) = 20 \). We are looking for \( n \).

\[ E(X) = n \times p \]

\[ n = \frac{E(X)}{p} = \frac{24}{0.6} = 40 \]

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) 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 \( p = 0.68 \) and \( n = 2 \). We want to have two successes out of two: \( P(X=2) = 0.6827 \times 0.6827 = 0.466 \)

b) We must find the inverse normal. With a calculator, we find that you have to be above $5537.86.

c) We find the inverse normal on a calculator, so: \( d = 4809.39 \, \text{USD} \)

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

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) (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 \( q \):

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