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If a variable X is normally distributed with a standard deviation of 8, what is the mean value of X, if 0.3413 of the values of X is greater than 58?

User Rinti
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1 Answer

1 vote

Answer:

55

Step-by-step explanation:

The z score is a score used in statistics to determine how many standard deviations the raw score is above or below the mean.it is given by the formula:


z=(x-\mu)/(\sigma)\\ Where\ \mu \ is\ the\ mean, x\ is\ the\ raw\ score\ and\ \sigma\ is\ the\ standard\ deviation

Given that:

σ = 8, P(z > 58) = 0.3413, x > 58/ Hence:

P(z < 58) = 1 - P(z > 58) = 1 - 0.3413 = 0.6587

P(z < 58) = 0.6587

From the normal distribution table, P(z < 58) = 0.6587 corresponds with a z score of 0.41


z=(x-\mu)/(\sigma)\\ \\0.41=(58-\mu)/(8)\\\\Cross-multiplying:\\\\58-\mu=3.28\\\\\mu=58-3,28\\\\\mu=54.72

μ ≅ 55

User Zpinter
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