A botanist wishes to estimate the typical number of seeds for a certain fruit. She samples 54 specimens and counts the number of seeds in each. Use her sample results (mean = 81.1, standard deviation = 8.4) to find the 90% confidence interval for the number of seeds for the species. Enter your answer as an open-interval (i. e., parentheses) accurate to 3 decimal places.

90% confidence interval for the number of seeds for the species

(78.8073 ,83.3927)

Step-by-step explanation:

Step(i):-

Given sample size 'n' =54

Mean of the sample x⁻ = 81.1

standard deviation of the sample 'S' = 8.4

90% confidence interval for the number of seeds for the species

Step(ii):-

Degrees of freedom

ν = n-1 = 54-1 =53

(81.1- 2.2927 ,81.1 + 2.2927 )

(78.8073 ,83.3927)

Final -

90% confidence interval for the number of seeds for the species

(78.8073 ,83.3927)

$12. that is my answer.

a 270° rotation about point p says on e d g e n u it y (had to space it out because it says it's a misspelling so i couldn't submit it