Use a unique mode to bound p, then sketch a different distribution

The source accepts either: the probability that a chosen person is underweight remains constant throughout the sample, or whether a chosen person is underweight is independent of whether other people are underweight. The full binomial model needs both constant p and independent trials; this part asks for one assumption in context.

The strict bounds requested by the source interpret 2 as the unique mode. Then $P(X=1)<P(X=2)$, so $\binom51p(1-p)^4<\binom52p^2(1-p)^3$. A mode at 2 rules out p=0 or p=1; therefore $p>0$ and $1-p>0$. Cancel positive factors to obtain $1-p<2p$, hence $p>1/3$.

Also $P(X=2)>P(X=3)$, giving $\binom52p^2(1-p)^3>\binom53p^3(1-p)^2$. Cancel positive factors to get $1-p>p$, hence $p<1/2$. This is the distinct Teacher completion. Together $1/3<p<1/2$, as required. If 2 were merely one of possibly tied modes, the endpoints would be included: at p=1/3 modes 1 and 2 tie, and at p=1/2 modes 2 and 3 tie.

This part gives a different parameter p=0.25; do not carry the range from (ii) into it. Use Y1=binompdf(5,0.25,X) and its table. The exact masses are

Plot separate points/crosses at 0 through 5. The source labels the first five heights approximately 0.237,0.396,0.264,0.0879,0.0146. Its final label (5,0) is inaccurate: $P(X=5)=(0.25)^5=1/1024\approx0.000977>0$, consistent with the source calculator’s 9.8E−4. The native sketch corrects this positive final mass; the original printed plot and text remain in source evidence. The mode of this new distribution is 1.