The number X of days in the summer months that a construction crew are cannot work because of the weather has the probability distribution X P(x) 6 0.03 7 0.08 8 0.15 9 0.20 10 0.19 11 0.16 12. 0.10 13 0.07 14 0.02 A. Find the probability that no more than ten days will be lost next summer. B. Find the probability that from 88 to 1212 will lost next summer. C. Find the probability that no days at all will be lost next summer. D. Compute the mean and standard deviation of X. Interpret the mean in the context of the problem.

$a)$

$p( x \le10 ) = p(x=6) + p(x =7) + p(x=8) + p(x=9) + p(x=10) = 0.03 + 0.08 + 0.15 + 0.2 + 0.19 = 0.65$

$b)$

$p( 8 \le x \le 12) = p(x=8) + p(x=9) + p(x=10) + p(x=11) + p(x=12) = 0.15 + 0.2 + 0.19 + 0.16 + 0.1 = 0.8$

$c)$

The probability that no days will be lost next summer will be given by,

$p(x=0)= 1 - ( p(x=6) + p(x=7) + p(x=8) + p(x=9) + p(10) + p(11) + p(12)) = 1 - 1 = 0$

 Therefore, the probability that there will be no days lost next summer is zero. We can conclude that there will be some days lost next summer.

$d)$

Mean

$E(x)=\sum xp(x)=(6\times 0.03)+(7\times 0.08)+(8\times 0.15)+(9\times 0.2)+(10\times0.19)+(11\times0.16)+(12\times0.1)+(13\times0.07)+(14\times0.02)=9.79\approx 10$

Interpretation,

The expected number of days that will be lost next summer due to weather is approximately equal to 10 days.

Standard deviation.

We first determine the variance given by,

$var(x)=E(x^2)-(E(x))^2$

Now,

$E(X^2)=\sum x^2p(x)=(36\times 0.03)+(49\times 0.08)+(64\times 0.15)+(81\times 0.2)+(100\times0.19)+(121\times0.16)+(144\times0.1)+(169\times0.07)+(196\times0.02)=99.31$

So,

$var(x)=99.31-9.79^2=3.4659$

Standard deviation is,

$sd(x)=\sqrt{var (x)}=\sqrt{3.4659}=1.86169278$