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 (68-95-99.7%) to estimate the probability of a lion living less than 7.2 years

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

7

step-by-step explanation:

i'm assuming you need to figure out how many campers go into each tent.

i got the answer by:

24+4=28

28/4=7