Information on a packet of seeds claims that the germination

Information on a packet of seeds claims that the germination rate is 80%. The package contains 100 seeds. (1.5 points each)

(a) If is the proportion of seeds that will germinate, what are the requirements for the sampling distribution of to be approximately normal? Are those requirements satisfied by our experimental setting? Explain your reasoning. What is the mean, standard deviation and sampling distribution of ?

(b) Calculate .

(c) Calculate .

(d) Explain the result in part (c) in such a way that someone with no training in statistics will understand.

Solution

There are totally 100 seeds and chances to geminate = p = 0.8

q = chance not to germinate = 0.2

As there are only two outcomes we can take this as Binomial

Mean = np = 8 and variance = 1.6

As sample size >100, X can be approximated to normal with mean 8 and std dev = 1.265

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Though binomial is discrete we do continuity correction to approximate to normal distribution

For example no of reading when a die is tossed,

outcomes are discrete as 1,2,3,4,5,6

When approximating to normal we write 1 as

-inftyx<1.5

1.5<x<2.5

2.5<x<3.5

...

...

5.5<x<6.5

and X>6,5

Like this we make it continuous and mean = np , std dev = rt of npq