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Paraphin [41]
3 years ago
6

Which of the following probabilities is the greatest for a standard normal distribution?P (negative 1.5 less-than-or-equal-to z

less-than-or-equal-to negative 0.5) P (negative 0.5 less-than-or-equal-to z less-than-or-equal-to 0.5) P (0.5 less-than-or-equal-to z less-than-or-equal-to 1.5) P (1.5 less-than-or-equal-to z less-than-or-equal-to 2.5)
Mathematics
2 answers:
Darina [25.2K]3 years ago
5 0

Answer:

the answer is below

Step-by-step explanation:

Which of the following probabilities is the greatest for a standard normal distribution?P (negative 1.5 less-than-or-equal-to z less-than-or-equal-to negative 0.5) P (negative 0.5 less-than-or-equal-to z less-than-or-equal-to 0.5) P (0.5 less-than-or-equal-to z less-than-or-equal-to 1.5) P (1.5 less-than-or-equal-to z less-than-or-equal-to 2.5)

Alinara [238K]3 years ago
4 0

Answer:

negative 0.5 less-than-or-equal-to z less-than-or-equal-to 0.5

Step-by-step explanation:

its the peak of the graph

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In a bag of m&amp;m's there are 5 brown 6 yellow 4 blue 3 green and 2 orange. What's the probability of getting 3 yellow m&amp;m
olasank [31]
There are 5+6+4+3+2=20 m&m's in the bag.
Calculate in how many ways you can choose 3 m&m's from 20:
_{20} C _3=\frac{20!}{3!(20-3)!}=\frac{20!}{3! \times 17!}=\frac{17! \times 18 \times 19 \times 20}{6 \times 17!}=\frac{18 \times 19 \times 20}{6}=3 \times 19 \times 20= \\&#10;=1140

There are 6 yellow m&m's.
Calculate in how many ways you can choose 3 m&m's from 6:
_6 C _3 = \frac{6!}{3!(6-3)!}=\frac{6!}{3! \times 3!}=\frac{3! \times 4 \times 5 \times 6}{3! \times 6}=\frac{4 \times 5 \times 6}{6}=4 \times 5=20

The probability is the number of ways of choosing 3 m&m's from 6 m&m's divided by the number of ways of choosing 3 m&m's from 20 m&m's.
P=&#10;\frac{20}{1140}=\frac{20 \div 20}{1140 \div 20}=\frac{1}{57}

The probability is 1/57.
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