What is 5 5/7 * 5 5/7
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a. percent of pig pregnancies that are longer than 106 days
Since we have a normal distribution here and the average number of days is 106, we can say that 50% of the pig pregnancies are longer than 106 days.
b. percent of pig pregnancies that are shorter than 111 days
To get the percentage, we will have to convert x = 111 days into a z-score first. The formula is:
where x = raw data, μ = population mean, and σ = population SD.
Since these 3 pieces of information are already given in the question, let's plug them into the equation above.
Therefore, 111 days is located 1 standard deviation to the right of the mean.
To find the percentage of pig pregnancies shorter than 111 days, we need to find the area covered to the left of 1 SD.
To find the area covered to the left of 1SD, we need to use the standard normal distribution table.
Based on the table, the area covered to the left of 1SD is 0.8413. Multiplying the area by 100, we get 84.13. Therefore, 84.13% of the pig pregnancies are shorter than 111 days.
