Martian potatoes begin to sprout very quickly after planting. Suppose X is the number of days after planting until a Martian potato sprouts. Then X has the following probability density function: f(x)={27e−x+314e−x2+114e−x4,0&lt;x&lt;∞0otherwise a) What is the probability that X is within 1 standard deviation of its expected value? b) What is the probability that X=.6?

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Martian potatoes begin to sprout very quickly after planting. Suppose X is the number of days after planting until a Martian potato sprouts. Then X has the following probability density function:  f(x)={27e−x+314e−x2+114e−x4,0&amp;lt;x&amp;lt;∞0otherwise  a) What is the probability that X is within 1 standard deviation of its expected value? b) What is the probability that X=.6?

Novembruuo · Accepted Answer

Step 1 a) From the given information, the probability density function for X is, f(x)={27e−x+314e−x2+114e−x4,0&amp;lt;x&amp;lt;∞0otherwise  Step 2 Consider, E(X)=∫0∞xf(x)dx =∫0∞x(27e−x+314e−x2+114e−x4)dx =∫0∞(27xe−x+314xe−x2+114xe−x4)dx =167 E(X2)=∫0∞x2f(x)dx =∫0∞x2(27e−x+314e−x2+114e−x4)dx =∫0∞(27x2e−x+314x2e−x2+114x2e−x4)dx =47+247+647 =927 Step 3 Therefore, SD(X)=E(X2)−[E(X)]2 =927−(167)2 =2.8139 The probability that X is within 1 standard deviation of its expected value is 0.8810 and it is calculated below:

Martian potatoes begin to sprout very quickly after planting. Suppose

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