Management at a seaside resort is publishing a brochure and wants to include a statement about the proportion of clear days during their peak season. out of a random sample of 150 days from over the last two peak seasons, 117 days were recorded as clear. they want to estimate the proportion of clear days to within a 5% margin of error with a 95% confidence interval. what's the sample size necessary to construct this interval? a. 384b. 264c. 383d. 385e. 263

B. 264

Step-by-step explanation:

Formula to calculate the sample size 'n' , if the prior estimate of the population proportion (p) is available:

, where z = Critical z-value corresponds to the given confidence interval

E=  margin of error

Let p be the population proportion of clear days.

As per given , we have

Prior sample size : n= 150

Number of clear days in that sample = 117

Prior estimate of the population proportion of clear days =

E= 0.05

The critical z-value corresponding to 95% confidence interval = z*= 1.95 (By z-table)

Then, the required sample size will be :

Simplify ,

Hence, the sample size necessary to construct this interval =264

Thus the correct option is B. 264

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.