Answer: 0.0129
Step-by-step explanation:
Given : Sample size : n=32
The average amount of time spent texting over a one-month period is :
Standard deviation :
We assume that the time spent texting over a one-month period is normally distributed.
z-score :
For x= 199
Now by using standard normal table, the probability that the average amount of time spent using text messages is more than 199 minutes will be :-
Hence, the required probability = 0.0129
Answer:
Is that thick line an equal or a minis sign? I cant see it.
Answer: a. n= 1068
b. n= 164
Step-by-step explanation:
The formula to find the sample size :
, where p=prior population proportion , z* = critical z-value and E = Margin of error.
Here , let p=proportion of computers that use a new operating system.
Given : Confidence level = 95%
i.e. z* = 1.96 [by z-table]
Margin of error : E = 3% =0.03
a. If p is unknown , then we assume p=0.5
Then,
i.e. n= 1068
b. p=0.96
Then,
i.e. n= 164.