If x is a binomial random variable with n =8 and p =0.6, the mean value of x is

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We start by identifying the key parameters of the binomial random variable: n = 8 trials and success probability p = 0.6. For a binomial distribution, the mean (expected value) is given by μ = np.

Option 1: 2.2. This value would correspond to np = 2.2, which would require p to be 2.2/8 = 0.275 if n = 8, or a different n. Since p is 0.6, 2.2 does not match the formula μ = np, so this is incorrect.

Option 2: 4.8. Calculating np yields 8 × 0.6 = 4.8. This aligns exactly with the mean of a binomial distribution, so this option matches the theoretical formula.

Option 3: 4. This would imply np = 4, which could occur if p = 0.5 (since 8 × 0.5 = 4) but here p = 0.6, so this value does not reflect the given parameters. It looks like a miscalculation if one assumes p = 0.5 or misapplies the formula.

Option 4: 1. This is far from np = 4.8 and would require either a much smaller p or a much smaller n. With n = 8 and p = 0.6, the mean cannot be as low as 1, so this option is not consistent with the binomial mean formula.

In summary, the only option that satisfies the mean formula μ = np with the given n and p is 4.8, since 8 × 0.6 equals 4.8. The other values fail to satisfy the fundamental relationship between the mean, n, and p for a binomial distribution.

If x is a binomial random variable with n =8 and p =0.6, the mean value of x is

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