"A sample of 20 randomly chosen water melons was taken from a large population, and their weights were measured. The mean weight of the sample was 105 lb. and the standard deviation was 15 lb. Calculate (correct to one decimal place) 99.5% confidence limits for the mean weight of the whole population of watermelons."

(97.98, 112.020)

Step-by-step explanation: We are going to construct a 95% confidence level interval for mean weight of melon.

From the parameters given to us, we conclude that the critical value of the interval will be gotten using a t distribution table, this is because sample size is less than 30 ( that's 20) and we are given the sample standard deviation (s =15lb)

Our parameters are given as follows

Sample mean =x = 105lb

Sample standard deviation =s= 15lb

Sample size = n = 20

To construct the 95% confidence interval, it denotes that we have a 5% level of significance.

The confidence interval formulae is given below as

u = x + tα/2 × s/√n For the upper limit

u = x - tα/2 × s/√n For the lower limit.

tα/2 = t critical value of the test ( which can be gotten using a t distribution table).

tα/2 is gotten by checking the value of degree of freedom (which is sample size - 1) against the level of significance for a 2 tailed test (α/2 = 0.025%) on a t distribution table

For upper limit, we have that

u = 105 + 2.093×15/√20

u = 105 + 2.093× (3.3541)

u = 105 + 7.020

u = 112.020.

For lower limit, we have that

u = 105 - 2.093×15/√20

u = 105 - 2.093× (3.3541)

u = 105 - 7.020

u = 97.98

Confidence interval (97.98, 112.020)

$8

step-by-step explanation:

it is pretty much backwards.

you take the 4x2 multiply that, and that is = $8

walter needs to make at least 5 more cookies but no more than 40. 5 ≤ x ≤ 40

step-by-step explanation:

20 ≤ x + 15 ≤ 55

subtract 15 from all parts

20-15 ≤ x + 15-15 ≤ 55-15

5 ≤ x ≤ 40

this means how many more cookies he has to make

walter needs to make at least 5 more cookies, but no more than 40 cookies; 5 ≤ x ≤ 40