Given:
100 students:
pass = + 4
fail =+ 0
90 students passed: 90 * 4 = 360
10 students failed: 10 * 0 = 0
mean: 360 / 100 = 36
standard deviation:
36 - 4 = 32 ⇒ 32² = 1,024 * 90 = 92,160
36 - 0 = 36 ⇒ 36² = 1,296 * 10 = 12,960
92,160 + 12,960 = 105,120
105,120 / 100 = 1,051.20 ⇒ variance
Standard deviation = √1,051.20 = 32.42
Answer:
B. z Subscript alpha divided by 2 zα/2 = 1.96.
Step-by-step explanation:
We are given that we want to construct a confidence interval. For this, the summary statistics for randomly selected weights of newborn girls:
n = 236,
= 30.3 hg, s = 7.2 hg. The confidence level is 95%.
As we can clearly see here that the population standard deviation is unknown and the sample size is also very large.
It has been stated that when the population standard deviation is unknown, we should use t-distribution but since the sample size is very large so we can use z distribution also as it is stated that at very large samples; the t-distribution corresponds to the z-distribution.
Here,
= level of significance = 1 - 0.95 = 0.05 or 5%
= 0.025 or 2.5%
So, the value of
in the z table is given as 1.96 with a 2.5% level of significance.
Your answer will be C. :)
We replace two point into any equation
there
y-4=-4/3(x+3)
right.