Atrucking firm has a large inventory of spare parts that have been in storage for a long time. it knowsthat some proportion of these spare parts will have deteriorated to the point where they are no longerusable. in order to estimate this proportion, the firm tests 75 parts, and finds that 0.25 of them are notusable. find the 95% confidence interval for the population proportion of parts that are not usable.
a. 0.25(1 0.25) 0.25(1 0.25) c 0.25 1.96 p 0.25 1.96 0.95 75 75
b. 0.25(1 0.25) 0.25(1 0.25) c 0.25 1.96 p 0.25 1.96 0.95 75 75
c. 0.25(1 0.25) 0.25(1 0.25) c 0.25 1.645 p 0.25 1.645 0.95 75 75
d. 0.25(1 0.25) 0.25(1 0.25) c 0.25 1.645 p 0.25 1.645 0.95 75 75

Step-by-step explanation:

The firm tests 75 parts, and finds that 0.25 of them are notusable

n = 75

x = 0.25 \times 75 = 18.75≈19

Confidence level = 95%

So, Z_\alpha at 95% = 1.96

Formula of confidence interval of one sample proportion:

Confidence interval

what

answer: first option i.e. 'a' is correct.

step-by-step explanation:

since we have given that

number of minutes to wrap a small gift box = 4 minutes

we need to find the number of gifts to be wrapped in 5 minutes

since there is a direct variation, as the number of minutes will increase with   number of gifts to be wrapped.

so, let the number of gifts to be wrapped in 5 minutes be x.

so, it becomes,

hence, first option 'a'   is correct.