Ammeters produced by a manufacturer are marketed under the specification that the standard deviation of gauge readings is no larger than 0.2 amp. One of these ammeters was used to make ten independent readings on a test circuit with constant current. The sample variance S 2 of these ten measurements is 0.0653 and it is reasonable to assume that the readings are normally distributed. Find the approximate probability that the sample variance will exceed 0.0653 if the true population variance is 0.04

0.101

Step-by-step explanation:

Given that :

Standard deviation, s = 0.2

Sample variance, s² = 0.0653

Sample size, n = 10

Population variance = σ² = 0.04

We use the Chi square distribution :

(n - 1)s² / σ²

For P(s² ≥ 0.065)

(n - 1)s² / σ² = (10 - 1)*0.0653 / 0.04

(n - 1)s² / σ² = 0.585 / 0.04 = 14.625

P(s² ≥ 14.625) = 0.101 ( chi squee calculator).

7 weeks

step-by-step explanation:

convert each fraction to the least common multiple of 20, so 3/4 = 15/20 & 1/10 = 2/20, then divide 15 by 2 and you get a little over 7, so the food will last 7 weeks.

i think this could be a possible answer (1 / 84) -1

good luck! : d