the lifespan of lions in a particular zoo are normally distributed. the average lion lives 10 years. the standard deviation is 1.4 years. use the empirical rule to estimate the probability of a lion living longer than 7.2 years
the lifespan of lions in a particular zoo are normally distributed. the average lion lives 10 years. the standard deviation is 1.4 years. use the empirical rule to estimate the probability of a lion living longer than 7.2 years
From the empirical rule we know that we have within one deviation from the mean 68% of the values, within two deviations 95% and within 3 deviations 99.7%. We want to find the following probability:
[tex]P( X<7.2)[/tex]
We can use the z score formula given by:
[tex]z = \frac{X-\mu}{\sigma}[/tex]
And replacing we got:
[tex]z = \frac{7.2-10}{1.4}= -2[/tex]
So we want to find the probability that the data lies below 2 deviations from the mean and using the empirical rule we got:
[tex](1-0.95)/2 = 0.025[/tex]
Step-by-step explanation:
For this problem we know that the average lion lives with a mean of [tex]\mu =10 years[/tex] and the deviation is [tex]\sigma =1.4[/tex]
From the empirical rule we know that we have within one deviation from the mean 68% of the values, within two deviations 95% and within 3 deviations 99.7%. We want to find the following probability:
[tex]P( X<7.2)[/tex]
We can use the z score formula given by:
[tex]z = \frac{X-\mu}{\sigma}[/tex]
And replacing we got:
[tex]z = \frac{7.2-10}{1.4}= -2[/tex]
So we want to find the probability that the data lies below 2 deviations from the mean and using the empirical rule we got:
[tex](1-0.95)/2 = 0.025[/tex]
the lion will only live 8.5 years
Step-by-step explanation:
16%
Step-by-step explanation:
2.5% probability of a lion living less than 7.2 years
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 10
Standard deviation = 1.4
Probability of a lion living less than 7.2 years
The normal distribution is symmetric, which means that 50% of the measures are above the mean and 50% are below.
7.2 = 10 - 2*1.4
So 7.2 is two standard deviations below the mean.
By the empirical rule, of those 50% of the measures below the mean, 95% is within 2 standard deviations of the mean, that is, between 7.2 and 10, and 5% are less than 7.2. So
p = 0.05*0.5 = 0.025
2.5% probability of a lion living less than 7.2 years
What is the actual question?