Answer:
Probability that fewer than three of them will have type O negative blood: P(x<3)=0.6047
Probability that more than 4 of them have type O negative blood: P(x>4)=0.0472
Probability that none of the donors had type O negative blood: P(x=0)=0.2342
Mean of the number of donors who have type O negative blood for a group this size: E=1
Standard deviation of the number of donors who have type O negative blood for a group this size: SD=2
Explanation:
These experiments behave as binomial distribution:
P(x)=
![(n!)/(x!(n-x)!). p^x.(1-p)^(n-x)](https://img.qammunity.org/2020/formulas/mathematics/college/xa96x7umadncf3loqriz01ajxg6u67rmb2.png)
where:
x: donors have type O negative blood
p: success outcome (0.07)
n: unrelated donors (20)
Fewer than three donors will have type O negative blood:
P(x<3)=P(x=0)+P(x=1)+P(x=2)=
+
+
![(20!)/(2!(20-2)!). (0.07)^2.(1-0.07)^(20-2)}](https://img.qammunity.org/2020/formulas/mathematics/college/xqdgzxka97jn8i8x28z9fu4fasjnxjgvao.png)
P(x<3)=0.2342+0.3526+0.2521=0.8389
More than 4 of donors have type O negative blood:
P(x>4)=1-P(x<4)= 1 - P(x=0) - P(x=1) - P(x=2) - P(x=3)= 1 - 0.2342 - 0.3526 - 0.2521 - 0.1139=0.0472
None of the donors had type O negative blood:
P(x=0)=
=0.2342
Mean of the number of donors who have type O negative blood:
E=n×p=20×0.07=1.4≅1
The mean number of donors that have type O negative blood is 1.
Standard deviation of the number of donors who have type O negative blood:
SD=
=
=1.5≅2
The standard deviation of the distribution of donors that have type O negative blood is 2.