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The amount of tea leaves in a can from a production line is normally distributed with = 110 grams and = 25 grams. A sample of 25 cans is to be selected. a. What is the probability that the sample mean will be between 100 and 120 grams? b. 95% of all sample means will be greater than how many grams? hwteazbjai. join her​

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95% of all sample means will be greater than 69 grams

Z score

Z score is used to determine the number of standard deviations the raw score is above or below the mean. It is given by:

x = raw score, μ = mean, σ = standard deviation, hence:

z = (x - μ)/σ

μ = 110, σ = 25

For x = 100:

z = (100 - 110) / 25 = -0.4

For x = 120:

z = (120 - 110) / 25 = 0.4

P(-0.4 < z < 0.4) = P(z < 0.4) - P(z < -0.4) = 0.6554 - 0.3446 = 31.08%

P(z > z*) = 0.95

1 - P(z < z*) = 0.95

P(z < z*) = 0.05

z* = -1.65

-1.64 = (x - 110)/25

x =69

95% of all sample means will be greater than 69 grams.

Find out more on Z score at: https://brainly.com/question/25638875

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