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Harman [31]
3 years ago
14

1. A bag contains rubber bands with lengths that are normally distributed with a mean of 6 cm of length, and a standard deviatio

n of 1.5 cm. What percentage of rubber bands can be expected to be shorter than 3 cm?
Mathematics
1 answer:
romanna [79]3 years ago
4 0
Roughly 1.7 percent of the bands are shorter than 3cm. We calculate the z score of the data point in standard distribution. By definition of z score, we use score minus mean divided by standard deviation. z=(3-6)/1.5=-2. A z score of -2 corresponds to approximately 1.7%, in other words, roughly 1.7 percent of data is less than 3cm.
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[6+4=10 points] Problem 2. Suppose that there are k people in a party with the following PMF: • k = 5 with probability 1 4 • k =
kirza4 [7]

Answer:

1). 0.903547

2). 0.275617

Step-by-step explanation:

It is given :

K people in a party with the following :

i). k = 5 with the probability of $\frac{1}{4}$

ii). k = 10 with the probability of $\frac{1}{4}$

iii). k = 10 with the probability $\frac{1}{2}$

So the probability of at least two person out of the 'n' born people in same month is  = 1 - P (none of the n born in the same month)

= 1 - P (choosing the n different months out of 365 days) = 1-\frac{_{n}^{12}\textrm{P}}{12^2}

1). Hence P(at least 2 born in the same month)=P(k=5 and at least 2 born in the same month)+P(k=10 and at least 2 born in the same month)+P(k=15 and at least 2 born in the same month)

= \frac{1}{4}\times (1-\frac{_{5}^{12}\textrm{P}}{12^5})+\frac{1}{4}\times (1-\frac{_{10}^{12}\textrm{P}}{12^{10}})+\frac{1}{2}\times (1-\frac{_{15}^{12}\textrm{P}}{12^{15}})

= 0.25 \times 0.618056 + 0.25 \times 0.996132 + 0.5 \times 1

= 0.903547

2).P( k = 10|at least 2 share their birthday in same month)

=P(k=10 and at least 2 born in the same month)/P(at least 2 share their birthday in same month)

= $0.25 \times \frac{0.996132}{0.903547}$

= 0.0.275617

6 0
2 years ago
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