Answer:
Yes we can conclude.
Step-by-step explanation:
The sampling distribution of can be approximated as a Normal Distribution only if:
np and nq are both equal to or greater than 10. i.e.
Both of these conditions must be met in order to approximate the sampling distribution of as Normal Distribution.
From the given data:
n = 50
p = 0.80
q = 1 - p = 1 - 0.80 = 0.20
np = 50(0.80) = 40
nq = 50(0.20) = 10
This means the conditions that np and nq must be equal to or greater than 10 is being satisfied. So, we can conclude that the sampling distribution of pˆ is approximately a normal distribution
Answer: 0.0170
Step-by-step explanation:
Given : The mean amount purchased by a typical customer at Churchill’s Grocery Store is $23.50, with a standard deviation of $5.00.
i.e.
We assume the distribution of amounts purchased follows the normal distribution.
Sample size : n=50
Let be the sample mean.
Formula :
Then, the probability that the sample mean is at least $25.00 will be :-
Hence, the likelihood the sample mean is at least $25.00= 0.0170
Answer:
Find the sixth term of the geometric sequence when a4=−8 and r=0.5
Step-by-step explanation: