Could you please check my work? Scenario: A study found that citizens spend on average $1950 per year on groceries with a standard deviation of $400. Assume that the variable is normally distributed. --> Find the probability that a sample of 50 citizens will have a mean less than $2000. z-score = (2000-1950)/[(400)/(√50)] = 50/56.569 = 0.88 probability = 0.3106 (according to z-table) 0.5 - 0.3106 = 0.1894 or 18.94% (I'm always confused about whether I should add or subtract from the 0.5, as I'm dealing with a half-distribution z-table)

  p(mean < 2000) ≈ 0.81

Step-by-step explanation:

Your table gives the probability the value is between z=0 and z=0.88. You want the probability the value is between -∞ and 0.88, so you have to add the probability it is between -∞ and zero. You must add 0.5 to the table value.

  p(Z < 0.88) = 0.5 +0.3106 = 0.8106

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Comment on table values

You probably need to do some interpolation of your table values to get accuracy to 4 significant digits. All of the calculators I use give the probability for a Z-score of 0.88388 to be about 0.8116, not 0.8106.

When working with a "half" table, you need to be aware of what the table is giving you and what you're trying to use it for. A quick sketch of the problem may be helpful. (see below)

i think it is 18

step-by-step explanation:

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step-by-step explanation: