Question

# Assume that farm sizes in a particular region are normally distributed with a population standard deviation...

Assume that farm sizes in a particular region are normally distributed with a population standard deviation of 150acres. A random sample of 50 farm sizes in this region is given below in acres. Estimate the mean farm size for this region with 90% confidence. Round your answers to two decimal places and use ascending order.

Size
202
1092
1321
319
331
938
73
200
200
229
160
486
370
256
351
770
170
288
177
165
67
194
353
296
293
2179
934
1490
107
79
1756
206
171
1312
188
438
474
131
56
204
1397
162
537
608
171
181
409
175
210
2612

Size
202
1092
1321
319
331
938
73
200
200
229
160
486
370
256
351
770
170
288
177
165
67
194
353
296
293
2179
934
1490
107
79
1756
206
171
1312
188
438
474
131
56
204
1397
162
537
608
171
181
409
175
210
2612

Solution:

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± Z*σ/sqrt(n)

From given data, we have

Xbar = 509.76

σ = 150

n = 50

Confidence level = 90%

Critical Z value = 1.6449

(by using z-table)

Confidence interval = Xbar ± Z*σ/sqrt(n)

Confidence interval = 509.76 ± 1.6449*150/sqrt(50)

Confidence interval = 509.76 ± 1.6449*21.2132

Confidence interval = 509.76 ± 34.8926

Lower limit = 509.76 - 34.8926 = 474.87

Upper limit = 509.76 + 34.8926 = 544.65

Confidence interval = (474.87, 544.65)

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