Respuesta :
Answer:
(a) [tex]X\sim N(\mu = 73, \sigma = 16)[/tex]
(b) 0.7910
(c) 0.0401
(d) 0.6464
Step-by-step explanation:
Let X = amount of time that people spend at Grover Hot Springs.
The random variable X is normally distributed with a mean of 73 minutes and a standard deviation of 16 minutes.
(a)
The distribution of the random variable X is:
[tex]X\sim N(\mu = 73, \sigma = 16)[/tex]
(b)
Compute the probability that a randomly selected person at the hot springs stays longer than 60 minutes as follows:
[tex]P(X>60)=P(\frac{X-\mu}{\sigma}>\frac{60-73}{16})\\=P(Z>-0.8125)\\=P(Z<0.8125)\\=0.7910[/tex]
*Use a z-table for the probability.
Thus, the probability that a randomly selected person at the hot springs stays longer than an hour is 0.7910.
(c)
Compute the probability that a randomly selected person at the hot springs stays less than 45 minutes as follows:
[tex]P(X<45)=P(\frac{X-\mu}{\sigma}<\frac{45-73}{16})\\=P(Z<-1.75)\\=1-P(Z<-1.75)\\=1-0.9599\\=0.0401[/tex]
*Use a z-table for the probability.
Thus, the probability that a randomly selected person at the hot springs stays less than 45 minutes is 0.0401.
(d)
Compute the probability that a randomly person spends between 60 and 90 minutes at the hot springs as follows:
[tex]P(60<X<90)=P(X<90)-P(X<60)\\=P(\frac{X-\mu}{\sigma}<\frac{90-73}{16})-P(\frac{X-\mu}{\sigma}<\frac{60-73}{16})\\=P(Z<1.0625)-P(Z<-0.8125)\\=0.8554-0.2090\\=0.6464[/tex]
*Use a z-table for the probability.
Thus, the probability that a randomly person spends between 60 and 90 minutes at the hot springs is 0.6464
