kindly check if there is mistakes in calculations.

givin mean ( / ) = 2200 Standard deviation ( 6 ) = 300 and n= 50 for 99 . confidence 2 scoocz 2. 576 CI = MIZO = 2200+ 2: 576 30 0 50 = 2200 I 109. 290 CI interval is ( 2090 . 709 , 2309. 290) (2 ) given mean ( M ) = 180 Standard deviation (6 ) = 30 and n = 100 (1) for 909%% contidence interval Z = 1- 645 CI = M+ ZO = 180 + 1.645 30 V10o = 180 I 4 . 935 CI interval is ( 175. 065, 184. 935 ) cs Scanned with CamScanner( 1i) for 95% confidence interval 2 = 1.960 CI = 1807 10960x 30 CI = 180 7 5. 880 C I interval is ( 174. 12 , 185: 88 ) ( in) for 99% confidence interval 2 = 2- 576 CI = 180 + 2 . 576 30 VIDO CI = 180 7 7072 8 CI interval is ( 172. 272, 187. 728 ) cs Scanned with CamScannerActivity 1: Estimation of Parameters Solve the following problem. 1. A nutritionist is interested in monitoring the calorie intake of women. It was found that in a selected random sample of 50 female adults, the average daily intake of meat products is 2,200 calories per day with a sample standard deviation of 300. Construct a 99% confidence interval of the mean intake of calories from meat products of female adults. 2. Construct a 90%, 95%, and 99% confidence interval in the mean weight of fresh oranges if 100 randomly selected fresh oranges have a mean weight of 180 grams (9) and a sample standard deviation of 30 g