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enter image description here $(x)$ is for number of complaints in a month. $P(X=x)$ is the probability

So $0.16$ probability for $0$ complaints a month. $0.24$ probability for $1$ complaint a month and so on.

The question is: What is the probability that the insurance company will receive 15 complaints in a year?

The solution: enter image description here

15/12 is for 15 complaint/12 months, 1.136 is the standard deviation, 1.1 is the expected value,

I dont understand how p(zā‰¤0.457) = 0.67. How do we get that number?

mangekyou
  • 25
  • From a normal distribution table. ā€“ zipirovich May 18 '20 at 17:07
  • Can I calculate it with Casio, or an online tool? ā€“ mangekyou May 19 '20 at 13:53
  • Online -- yes, of course you can. Search online for a "normal distribution table" or a "normal distribution calculator", whichever you prefer. As for calculators, such a Casio, I have no idea because I don't use them. Besides, asking about a "Casio" is not specific enough -- they manufacture a huge variety of calculators; and I presume their more advancded models can do this. ā€“ zipirovich May 19 '20 at 16:01

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