Answer: The 95% confidence interval is approximately (55.57, 58.43)
======================================================
Explanation:
At 95% confidence, the z critical value is about z = 1.960 which you find using a table or a calculator.
The sample size is n = 17
The sample mean is xbar = 57
The population standard deviation is sigma = 3
The lower bound of the confidence interval is
L = xbar - z*sigma/sqrt(n)
L = 57 - 1.960*3/sqrt(17)
L = 55.5738905247863
L = 55.57
The upper bound is
U = xbar + z*sigma/sqrt(n)
U = 57 + 1.960*3/sqrt(17)
U = 58.4261094752137
U = 58.43
Therefore the confidence interval (L, U) turns into (55.57, 58.43) which is approximate.
10.25 + 2.00 = 12.25...cost of pizza and soda
12.25 - 1.50x = 4.75....with x being the number of touchdowns
-1.50x = 4.75 - 12.25...I subtracted 12.25 from both sides
-1.50x = - 7.50...now divide both sides by -1.50
x = -7.50 / -1.50
x = 6......They would have to score 6 touchdowns
A^2+b^2=C^2
=
4+144=148
The square root of 148 is 12.165
round it down to 12
Hope this helps
Answer:
P(X is greater than 30) = 0.06
Step-by-step explanation:
Given that:
Sample proportion (p) = 0.5
Sample size = 30
The Binomial can be approximated to normal with:
To find:
P(X> 30)
So far we are approximating a discrete Binomial distribution using the continuous normal distribution. 30 lies between 29.5 and 30.5
Normal distribution:
x = 30.5, 
= 25, 
= 3.536
Using the z test statistics;
z = 1.555
The p-value for P(X>30) = P(Z > 1.555)
The p-value for P(X>30) = 1 - P (Z< 1.555)
From the z tables;
P(X> 30) = 1 - 0.9400
Thus;
P(X is greater than 30) = 0.06
and surely you know how much that is.
