Question

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A population has a mean of 300 and a standard deviation of 70. Suppose a sample of size 100 is selected and is used to estimate H. Use z-table.
a. What is the probability that the sample mean will be within +/- 6 of the population mean (to 4 decimals)? (Round z value in intermediate calculations to 2 decimal
places.)

Answer 1

Probability =0.9898

Explanation:

Given that, mean = 4 = 300 standard deviation = 70 n = 100 PT = H = 128 oT = 0 / V n = 70/ v 100 = 7 a) within 6 = 300 ±6 = 294, 306 P(294< ī < 306) = P[(294 - 300) 7 < (ĩ -H ï) / o ĩ < (306 300) /7)] = P(-0.86 < Z < 0.86) = P(Z < 0.86) - P(Z < -0.86) Using z table, 0.8051 - 0.1949 %3D %3D %3D = 0.6102 Probability = 0.6102 a) within 18 = 300 t 18 = 294, 318 P(294< ī < 306) = P[(282 - 300) / 7< (ã -H i) / oī< (318 - 300) /7)] = P(-2.57 < Z< 2.57) = P(Z < 2.57) - P(Z < -2.57) Using z table, =0.9949 - 0.0051 =0.9898 Probability =0.9898