The answer to your problem is 20%
Answer:
its c
Step-by-step explanation:
i took it
This problem is a combination of the Poisson distribution and binomial distribution.
First, we need to find the probability of a single student sending less than 6 messages in a day, i.e.
P(X<6)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)
=0.006738+0.033690+0.084224+0.140374+0.175467+0.175467
= 0.615961
For ALL 20 students to send less than 6 messages, the probability is
P=C(20,20)*0.615961^20*(1-0.615961)^0
=6.18101*10^(-5) or approximately
=0.00006181
The simplified form is C .
Answer:
Step-by-step explanation:
Given that:
Population Mean = 7.1
sample size = 24
Sample mean = 7.3
Standard deviation = 1.0
Level of significance = 0.025
The null hypothesis:

The alternative hypothesis:

This test is right-tailed.

Rejection region: at ∝ = 0.025 and df of 23, the critical value of the right-tailed test 
The test statistics can be computed as:



t = 0.980
Decision rule:
Since the calculated value of t is lesser than, i.e t = 0.980 <
, then we do not reject the null hypothesis.
Conclusion:
We conclude that there is insufficient evidence to claim that the population mean is greater than 7.1 at 0.025 level of significance.