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 \( q \):

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

Detailed Answer

a) The sum of the probabilities of all options should give 1. If we add up what's given in the table, we get \(\frac{6}{7}\), so the remaining probability is for when we miss the target.

b) The definition of a fair game is that the expected value \(E(X) = 0\). The expected value is:

\[10 \cdot \frac{1}{7} + 5 \cdot \frac{3}{7} + 1 \cdot \frac{2}{7} + k \cdot \frac{1}{7} = 0 \]

\[k = -27 \]

Detailed Answer

a) More than 60 means the number of cars he encounters is larger than or equal to 61. Hence, \(Pr(\text{Cars} > 60) = Pr(\text{Cars} \ge 61) = \text{poissoncdf}(50, 60, 200) = 0.09\).

b) This now becomes a Binomial distribution, with a probability of 0.09, and for 7 days (the length of a week). Thus the expected number of days is:

\[ E(X) = np = 7 \cdot 0.09 = 0.65 \]

c) We know each day the mean is 50, so for 7 days the mean will be 350. Thus, we need:

\[ Pr(\text{Cars} \le 299) = \text{poissoncdf}(350, 0, 299) = 0.0029 \]

Detailed Answer

a) Each lap has a mean time of 45 seconds, thus 3 laps will have a mean time of 135 seconds, as we can simply add the means.

b) We cannot simply add the standard deviations of the individual runs, but we can add the variances. The variance is the standard deviation squared. Thus, the overall variance will be:

\[ 3^2 + 3^2 + 3^2 = 27 \]

Taking the square root, the overall standard deviation is:

\[ \sqrt{27} \approx 5.20 \text{ seconds} \]

c) We need to find \(Pr(\text{Time} < 39)\), which can be done with the GDC. Using the normal distribution calculator, we find:

\[ \text{NormalCdf}(0, 39, 45, 3) = 0.022 \]

Detailed Answer

a)

b) One standard deviation is 1 grade, so being 0.5 under the mean is half a standard deviation.

c) \( Pr(\text{Grade} > 7.5) = \text{NormCDF}(7.5, 100, 6, 1) = 0.067 \)

d) (i) \( Pr(6 < \text{Grade} < 8) = \text{NormCDF}(6, 8, 6, 1) = 0.477 \)

d) (ii) Now the question has turned into a binomial distribution, with a sample of 30, and probability of 0.477. The expected value of a binomial distribution is simple to calculate, as we need to multiply the number of trials, with the probability, so \( 30 \cdot 0.477 = 14.32 \), so approximately 14 students.

e) \( X = \text{InvNorm}(0.75, 6, 1) = 6.67 \)