Can someone help me with this

Mathematics

1 answer:

Inga [223]2 years ago

7 0

Answer:

95% = Within two standard deviations from the mean

99.7% = Within three standard deviations from the mean

68% = Within one standard deviation from the mean

Step-by-step explanation:

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Assume the population of human body temperatures has a mean of 98.6 degrees Fahrenheit as is commonly believed. Assume the stand

Dennis_Churaev [7]

The probability that the mean of a sample of 106 randomly selected humans is lower than 98.5°F is 4.85%

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

Z score is given as:

z = (raw score - mean) ÷ (standard deviation/√sample size)

Given mean of 98.6°F, standard deviation is 0.62°F, sample size = 106

For x < 98.5:

z = (98.5 - 98.6) ÷ (0.62÷√106) = -1.66

P(z < -1.66) = 0.0485

The probability that the mean of a sample of 106 randomly selected humans is lower than 98.5°F is 4.85%

Find out more on equation at: brainly.com/question/2972832

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4 0

2 years ago

What’s 5e+|-12+f|+g e=-3 f=4 g=-1

AleksAgata [21]

Answer:

-8

Step-by-step explanation:

5 times -3 is -15

absolute value of -12 plus 4 is 8

-15 plus 8 plus -1 is -8

8 0

3 years ago

the numbers 1,2,3,4, and 5 are written on slips of paper, and 2 slips are drawn at random one at a time without replacet. find t

Tresset [83]

Consider such events:

A - slip with number 3 is chosen;

B - the sum of numbers is 4.

You have to count $Pr(A|B).$

Use formula for conditional probability:

$Pr(A|B)=\dfrac{Pr(A\cap B)}{Pr(B)}.$

1. The event $A\cap B$ consists in selecting two slips, first is 3 and second should be 1, because the sum is 4. The number of favorable outcomes is exactly 1 and the number of all possible outcomes is 5·4=20 (you have 5 ways to select 1st slip and 4 ways to select 2nd slip). Then the probability of event $A\cap B$ is

$Pr(A\cap B)=\dfrac{1}{20}.$

2. The event $B$ consists in selecting two slips with the sum 4. The number of favorable outcomes is exactly 2 (1st slip 3 and 2nd slip 1 or 1st slip 1 and 2nd slip 3) and the number of all possible outcomes is 5·4=20 (you have 5 ways to select 1st slip and 4 ways to select 2nd slip). Then the probability of event $B$ is