Question
For a binomial distribution, standard deviation is $0.8$ and probability of failure is $\frac{2}{3}$. Find the mean of this distribution.
✓
Answer
Here, standard deviation of binomial distribution $=\sqrt{ \mathrm{npq}}$
$\sqrt{n p q}=0.8$
$\therefore \mathrm{npq}=0.64$
Probability 7 of failure $q=\frac{2}{3}$
Mean of the distribution:
Putting, $\mathrm{q}=\frac{2}{3}$ in $n p q=0.64$
$\mathrm{np} \times \frac{2}{3}=0.64$
$\therefore \mathrm{np}=\frac{0.64 \times 3}{2}=096$
Hence, the mean of the distribution obtained is $0.96 .$
$\sqrt{n p q}=0.8$
$\therefore \mathrm{npq}=0.64$
Probability 7 of failure $q=\frac{2}{3}$
Mean of the distribution:
Putting, $\mathrm{q}=\frac{2}{3}$ in $n p q=0.64$
$\mathrm{np} \times \frac{2}{3}=0.64$
$\therefore \mathrm{np}=\frac{0.64 \times 3}{2}=096$
Hence, the mean of the distribution obtained is $0.96 .$
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