Question

ANSWER IF YOU HAVE SOLUTION OR ELSE YOU WILL BE REPORTED!
40% of the employees at a large corporation are female. A sample of 50 employees is taken and gender is recorded. What is the probability of a sample proportion of females in the sample between
.35 and .45?
a) 0.0693 b) 0.2649
c) 0.5284 d) 0.72
Kindly explain it too with solution
will give you brainliest for correct answer

Answer 1

Answer:

c) 0.5284

Step-by-step explanation:

The z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by the formula:

$z=\frac{p_s-p}{\sigma_p} \\\\where\ p_s=sample\ proportion,p=population\ proportion,\\\sigma_p=standard \ error=\sqrt{ \frac{p(1-p)}{n}},n=sample\ size\\\\ z=\frac{p_s-p}{\sqrt{ \frac{p(1-p)}{n}}} \\\\Given \ that\ n=50,p=40\%=0.4\\\\For\ p_s=0.35:\\\\z=\frac{0.35-0.4}{\sqrt{\frac{0.4(1-0.4)}{50} } } =-0.72\\\\For\ p_s=0.45:\\\\z=\frac{0.45-0.4}{\sqrt{\frac{0.4(1-0.4)}{50} } } =0.72$

Hence from the normal distribution table, P(0.35 < x < 0.45) = P(-0.72 < z < 0.72) = P(z < 0.72) - P(z < -0.72) = 0.7642 - 0.2358 = 0.5284