Respuesta :
Answer:
The probability that at most 50 say that they drink coffee during exam week is 0.166.
Step-by-step explanation:
The random variable X can be defined as the number of college students who prefer to drink more coffee during the exams week.
The probability of the random variable X is p = 0.67.
A random sample of n = 80 college students are selected.
The response of every students is independent of the others.
The random variable X follows a Binomial distribution with parameters n = 80 and p = 0.67.
But the sample selected is too large and the probability of success is close to 0.50.
So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:
- np ≥ 10
- n(1 - p) ≥ 10
Check the conditions as follows:
[tex]np=80\times 0.67=53.6>10\\n(1-p)=80\times (1-0.67)=26.4>10[/tex]
Thus, a Normal approximation to binomial can be applied.
[tex]X\sim N(np, np(1-p))[/tex]
The mean of the distribution of X is:
[tex]\mu=np=80\times 0.67=53.6[/tex]
The standard deviation of the distribution of X is:
[tex]\sigma=\sqrt{np(1-p)}=\sqrt{80\times 0.67\times (1-0.67)}=4.206[/tex]
A Normal distribution is a continuous distribution. So, the probability at a point cannot be computed for the Normal distribution. To compute the probability at a point we need to apply the continuity correction.
Compute the probability that at most 50 say that they drink coffee during exam week as follows:
Apply continuity correction:
[tex]P(X\leq 50)=P(X<50-0.50)[/tex]
[tex]=P(X<49.50)\\\\=P(\frac{X-\mu}{\sigma}<\frac{49.50-53.6}{4.206})\\\\=P(Z<-0.97)\\\\=0.16602\\\\\approx 0.166[/tex]
*Use a z-table for the probability.
Thus, the probability that at most 50 say that they drink coffee during exam week is 0.166.
