STATS HELP!!!?

A particular fruit's weights are normally distributed, with a mean of 338 grams and a standard deviation of 15 grams.

If you pick 16 fruit at random, what is the probability that their mean weight will be between 340 grams and 347 grams

2 Answers

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  • 1 month ago

    Standard deviation of the sampling distribution of x = SEx = Sigma/sqrt n

    = 15/sqrt 16

    = 3.75

    z = (Xbar - Mu)/SEx

    z1 = (340 - 338)/3.75 = + 0.53 approximately

    z2 = (347 - 338)/3.75 = + 2.40

    Required probability = P(340 < Xbar < 347)

    = P(0.53 < z < 2.40)

    = P(z < 2.40) - P(z < 0.53)

    = 0.9918 - 0.7019

    = 0.2899

    z-score chart showing the area under the standard normal curve on left of z is used...

  • 1 month ago

    The standard error of the sampling distribution will be 15/sqrt(16) = 3.75.

    The z-scores for the given bounds are

    z = 2/3.75 and z = 9/3.75, or

    0.5333 < z < 2.4000.

    Go to a normal distribution to find what percent of data lie between these two z-scores.

    I get around 29%.

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