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Sagot :
Let's tackle the given problems step-by-step.
### Problem (a)
Given:
- Mean, [tex]\(\mu = 20\)[/tex]
- Standard Deviation, [tex]\(\sigma = 4\)[/tex]
#### Part (a)(i): Probability that [tex]\( X \leq 24 \)[/tex]
1. Normalize the variable [tex]\( X \)[/tex] to the standard normal variable [tex]\( Z \)[/tex]:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
2. Calculate the z-score for [tex]\( X = 24 \)[/tex]:
[tex]\[ Z_1 = \frac{24 - 20}{4} = 1.0 \][/tex]
3. Find the cumulative probability [tex]\( P(X \leq 24) \)[/tex] using the z-score value:
[tex]\[ P(X \leq 24) = P(Z \leq 1.0) = 0.8413447460685429 \][/tex]
Therefore, the probability that [tex]\( X \leq 24 \)[/tex] is approximately [tex]\( 0.8413 \)[/tex].
#### Part (a)(ii): Probability that [tex]\( 16 \leq X \leq 28 \)[/tex]
1. Calculate the z-scores for [tex]\( X = 16 \)[/tex] and [tex]\( X = 28 \)[/tex]:
[tex]\[ Z_2 = \frac{16 - 20}{4} = -1.0 \][/tex]
[tex]\[ Z_3 = \frac{28 - 20}{4} = 2.0 \][/tex]
2. Find the cumulative probabilities:
[tex]\[ P(X \leq 16) = P(Z \leq -1.0) = 0.15865525393145707 \][/tex]
[tex]\[ P(X \leq 28) = P(Z \leq 2.0) = 0.9772498680518208 \][/tex]
3. Calculate the probability for [tex]\( 16 \leq X \leq 28 \)[/tex]:
[tex]\[ P(16 \leq X \leq 28) = P(X \leq 28) - P(X \leq 16) \][/tex]
[tex]\[ P(16 \leq X \leq 28) = 0.9772498680518208 - 0.15865525393145707 = 0.8185946141203637 \][/tex]
Therefore, the probability that [tex]\( 16 \leq X \leq 28 \)[/tex] is approximately [tex]\( 0.8186 \)[/tex].
### Problem (b)
Given:
- A fair coin (probability of head, [tex]\( p = 0.5 \)[/tex])
- Number of tosses, [tex]\( n = 6 \)[/tex]
#### Part (b)(i): Probability of getting no head
1. Calculate the probability of getting 0 heads in 6 tosses:
[tex]\[ P(\text{no head}) = (1 - p)^n = (1 - 0.5)^6 = 0.5^6 = 0.015625 \][/tex]
Therefore, the probability of getting no head is [tex]\( 0.015625 \)[/tex].
#### Part (b)(ii): Probability of getting at least one head
1. Calculate the probability of getting at least one head:
[tex]\[ P(\text{at least one head}) = 1 - P(\text{no head}) \][/tex]
[tex]\[ P(\text{at least one head}) = 1 - 0.015625 = 0.984375 \][/tex]
Therefore, the probability of getting at least one head is [tex]\( 0.984375 \)[/tex].
In summary:
- [tex]\( P(X \leq 24) \approx 0.8413 \)[/tex]
- [tex]\( P(16 \leq X \leq 28) \approx 0.8186 \)[/tex]
- Probability of getting no head = [tex]\( 0.015625 \)[/tex]
- Probability of getting at least one head = [tex]\( 0.984375 \)[/tex]
### Problem (a)
Given:
- Mean, [tex]\(\mu = 20\)[/tex]
- Standard Deviation, [tex]\(\sigma = 4\)[/tex]
#### Part (a)(i): Probability that [tex]\( X \leq 24 \)[/tex]
1. Normalize the variable [tex]\( X \)[/tex] to the standard normal variable [tex]\( Z \)[/tex]:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
2. Calculate the z-score for [tex]\( X = 24 \)[/tex]:
[tex]\[ Z_1 = \frac{24 - 20}{4} = 1.0 \][/tex]
3. Find the cumulative probability [tex]\( P(X \leq 24) \)[/tex] using the z-score value:
[tex]\[ P(X \leq 24) = P(Z \leq 1.0) = 0.8413447460685429 \][/tex]
Therefore, the probability that [tex]\( X \leq 24 \)[/tex] is approximately [tex]\( 0.8413 \)[/tex].
#### Part (a)(ii): Probability that [tex]\( 16 \leq X \leq 28 \)[/tex]
1. Calculate the z-scores for [tex]\( X = 16 \)[/tex] and [tex]\( X = 28 \)[/tex]:
[tex]\[ Z_2 = \frac{16 - 20}{4} = -1.0 \][/tex]
[tex]\[ Z_3 = \frac{28 - 20}{4} = 2.0 \][/tex]
2. Find the cumulative probabilities:
[tex]\[ P(X \leq 16) = P(Z \leq -1.0) = 0.15865525393145707 \][/tex]
[tex]\[ P(X \leq 28) = P(Z \leq 2.0) = 0.9772498680518208 \][/tex]
3. Calculate the probability for [tex]\( 16 \leq X \leq 28 \)[/tex]:
[tex]\[ P(16 \leq X \leq 28) = P(X \leq 28) - P(X \leq 16) \][/tex]
[tex]\[ P(16 \leq X \leq 28) = 0.9772498680518208 - 0.15865525393145707 = 0.8185946141203637 \][/tex]
Therefore, the probability that [tex]\( 16 \leq X \leq 28 \)[/tex] is approximately [tex]\( 0.8186 \)[/tex].
### Problem (b)
Given:
- A fair coin (probability of head, [tex]\( p = 0.5 \)[/tex])
- Number of tosses, [tex]\( n = 6 \)[/tex]
#### Part (b)(i): Probability of getting no head
1. Calculate the probability of getting 0 heads in 6 tosses:
[tex]\[ P(\text{no head}) = (1 - p)^n = (1 - 0.5)^6 = 0.5^6 = 0.015625 \][/tex]
Therefore, the probability of getting no head is [tex]\( 0.015625 \)[/tex].
#### Part (b)(ii): Probability of getting at least one head
1. Calculate the probability of getting at least one head:
[tex]\[ P(\text{at least one head}) = 1 - P(\text{no head}) \][/tex]
[tex]\[ P(\text{at least one head}) = 1 - 0.015625 = 0.984375 \][/tex]
Therefore, the probability of getting at least one head is [tex]\( 0.984375 \)[/tex].
In summary:
- [tex]\( P(X \leq 24) \approx 0.8413 \)[/tex]
- [tex]\( P(16 \leq X \leq 28) \approx 0.8186 \)[/tex]
- Probability of getting no head = [tex]\( 0.015625 \)[/tex]
- Probability of getting at least one head = [tex]\( 0.984375 \)[/tex]
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