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A recent study found that the life expectancy of a people living in Africa is normally distributed with an average of 53 years with a standard deviation of 7.5 years. If a person in Africa is selected at random, what is the probability that the person will die before the age of 65?

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Answer:

P(X<65)=0.9452

Step-by-step explanation:

This is a normal distribution problem.

-Given the mean age is 53 years and the standard deviation is 75, the probability of dying before age 65 is calculated as:

[tex]z=\frac{\bar x-\mu}{\sigma}\\\\\\=\frac{65-53}{7.5}\\\\=1.60[/tex]

#We check the value of z=1.60 on the z table:

[tex]P(X<65)=0.5+0.4452\\\\=0.9452[/tex]

Hence, the probability of dying before 65 is 0.9452

the probability that the person will die before the age of 65 is 0.9452.

  • The calculation is as follows:

[tex]P(X<65 ) = P[(X- \mu ) \div \sigma < (65 -53) \div 7.5][/tex]

= P(z <1.6 )

Now here we Using z table

So,  

= 0.9452

Learn more: https://brainly.com/question/25914450?referrer=searchResults

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