Question

b. what is the probability that at most 7 of them have blue eyes?
10.5
c. what is the probability that less than 7 of them have blue eyes?
10.5
d. what is the probability that at least 7 of them have blue eyes?
0.895
e. what is the probability that more than 7 of them have blue eyes?
0.0103
f. compute the expected value.
0.49
g. compute the standard deviation.
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Explanation:

To solve these probability and statistics problems, we assume this is a binomial distribution problem (since we're dealing with the number of successes, here "having blue eyes"). Let's denote the number of trials as \( n \), the probability of success (having blue eyes) as \( p \), and the number of successes as \( X \).

Part b: Probability that at most 7 have blue eyes

"At most 7" means \( P(X \leq 7) \). For a binomial distribution, this is the sum of probabilities from \( X = 0 \) to \( X = 7 \). However, the given answer is \( 10.5 \), which seems incorrect (probabilities must be between 0 and 1). There may be a typo or missing context (e.g., \( n \) or \( p \) not provided).

Part c: Probability that less than 7 have blue eyes

"Less than 7" means \( P(X < 7) = P(X \leq 6) \). Again, the given answer \( 10.5 \) is invalid (probability > 1).

Part d: Probability that at least 7 have blue eyes

"At least 7" means \( P(X \geq 7) = 1 - P(X \leq 6) \). The given answer is \( 0.895 \), which is a valid probability (between 0 and 1).

Part e: Probability that more than 7 have blue eyes

"More than 7" means \( P(X > 7) = P(X \geq 8) \). The given answer \( 0.0103 \) is valid.

Part f: Expected value

For a binomial distribution, the expected value (mean) is \( \mu = np \). The given answer is \( 0.49 \), so \( np = 0.49 \).

Part g: Standard deviation

For a binomial distribution, the standard deviation is \( \sigma = \sqrt{np(1 - p)} \). To compute this, we need \( n \) and \( p \). From the expected value \( np = 0.49 \), let’s assume \( n = 1 \) (unlikely) or \( p = 0.49 \) and \( n = 1 \) (also unlikely). Alternatively, if \( n = 100 \) and \( p = 0.0049 \), but this is speculative. Without \( n \) or \( p \), we can’t verify, but the formula is \( \sigma = \sqrt{np(1 - p)} \).

Final Answers (as given, with caveats for validity):

b. \( \boldsymbol{10.5} \) (invalid, likely a typo)
c. \( \boldsymbol{10.5} \) (invalid, likely a typo)
d. \( \boldsymbol{0.895} \)
e. \( \boldsymbol{0.0103} \)
f. \( \boldsymbol{0.49} \)
g. (Depends on \( n \) and \( p \); formula: \( \sqrt{np(1 - p)} \))

Note: The answers for parts b and c are invalid because probabilities cannot exceed 1. This suggests missing context (e.g., \( n \) or \( p \)) or a typo.

Answer:

To solve these probability and statistics problems, we assume this is a binomial distribution problem (since we're dealing with the number of successes, here "having blue eyes"). Let's denote the number of trials as \( n \), the probability of success (having blue eyes) as \( p \), and the number of successes as \( X \).

Part b: Probability that at most 7 have blue eyes

"At most 7" means \( P(X \leq 7) \). For a binomial distribution, this is the sum of probabilities from \( X = 0 \) to \( X = 7 \). However, the given answer is \( 10.5 \), which seems incorrect (probabilities must be between 0 and 1). There may be a typo or missing context (e.g., \( n \) or \( p \) not provided).

Part c: Probability that less than 7 have blue eyes

"Less than 7" means \( P(X < 7) = P(X \leq 6) \). Again, the given answer \( 10.5 \) is invalid (probability > 1).

Part d: Probability that at least 7 have blue eyes

"At least 7" means \( P(X \geq 7) = 1 - P(X \leq 6) \). The given answer is \( 0.895 \), which is a valid probability (between 0 and 1).

Part e: Probability that more than 7 have blue eyes

"More than 7" means \( P(X > 7) = P(X \geq 8) \). The given answer \( 0.0103 \) is valid.

Part f: Expected value

For a binomial distribution, the expected value (mean) is \( \mu = np \). The given answer is \( 0.49 \), so \( np = 0.49 \).

Part g: Standard deviation

For a binomial distribution, the standard deviation is \( \sigma = \sqrt{np(1 - p)} \). To compute this, we need \( n \) and \( p \). From the expected value \( np = 0.49 \), let’s assume \( n = 1 \) (unlikely) or \( p = 0.49 \) and \( n = 1 \) (also unlikely). Alternatively, if \( n = 100 \) and \( p = 0.0049 \), but this is speculative. Without \( n \) or \( p \), we can’t verify, but the formula is \( \sigma = \sqrt{np(1 - p)} \).

Final Answers (as given, with caveats for validity):

b. \( \boldsymbol{10.5} \) (invalid, likely a typo)
c. \( \boldsymbol{10.5} \) (invalid, likely a typo)
d. \( \boldsymbol{0.895} \)
e. \( \boldsymbol{0.0103} \)
f. \( \boldsymbol{0.49} \)
g. (Depends on \( n \) and \( p \); formula: \( \sqrt{np(1 - p)} \))

Note: The answers for parts b and c are invalid because probabilities cannot exceed 1. This suggests missing context (e.g., \( n \) or \( p \)) or a typo.