The botanist would like to estimate the average weight of mustard seeds grown in a certain type of soil; weight of the seeds is known to follow a normal distribution. The botanist collects a random sample of 10 mustard seeds and finds that the sample average and sample standard deviation are 528 mg and 49 mg respectively. Select from the following the correct 90% confidence interval for the true average weight of mustard seeds grown in Soil Type A using the above sample data. a. (480.1, 539.9)
b. (525.2, 534.8)
c. (479.4, 540.6)
d. (473.6, 546.4)
e. The necessary assumptions have not been satisfied and the calculation cannot be performed.
f. (469.4, 550.6)

The correct 90% confidence interval for the true average weight of mustard seeds grown in Soil Type A using the above sample data is (499.6, 556.4).

Step-by-step explanation:

We have the standard deviation for the sample, so we use the t-distribution to solve this question.

T interval

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 10 - 1 = 9

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 9 degrees of freedom(y-axis) and a confidence level of . So we have T = 1.833

The margin of error is:

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 528 - 28.4 = 499.6 mg

The upper end of the interval is the sample mean added to M. So it is 528 + 28.4 = 556.4 mg

The correct 90% confidence interval for the true average weight of mustard seeds grown in Soil Type A using the above sample data is (499.6, 556.4).

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !

answer: rolling a number cube and flipping a coin.

step-by-step explanation: