question #7\nthe average number of candies per bag is estimated to be between 21 and 35. determine the…

question #7\nthe average number of candies per bag is estimated to be between 21 and 35. determine the sample mean.\n28\n30\n32\n60\nquestion #8\nfor a population that is distributed normally with a mean of 62 and a standard deviation of 15, calculate p(68 < x < 88).\n.3031\n.0310\n.7341\n.9690

question #7\nthe average number of candies per bag is estimated to be between 21 and 35. determine the sample mean.\n28\n30\n32\n60\nquestion #8\nfor a population that is distributed normally with a mean of 62 and a standard deviation of 15, calculate p(68 < x < 88).\n.3031\n.0310\n.7341\n.9690

Answer

Explanation:

Step1: Recall sample - mean formula for confidence interval

The sample mean for a confidence interval is the mid - point of the interval. The formula for the mid - point of an interval ([a,b]) is (\bar{x}=\frac{a + b}{2}), where (a = 21) and (b = 35).

Step2: Calculate the sample mean

(\bar{x}=\frac{21+35}{2}=\frac{56}{2}=28)

Step3: For normal distribution probability calculation

First, standardize the values using the formula (z=\frac{x-\mu}{\sigma}), where (\mu = 62) and (\sigma = 15). For (x = 68), (z_1=\frac{68 - 62}{15}=\frac{6}{15}=0.4) For (x = 88), (z_2=\frac{88 - 62}{15}=\frac{26}{15}\approx1.73)

Then, use the standard normal distribution table. (P(68<X<88)=P(0.4<Z<1.73)=P(Z < 1.73)-P(Z < 0.4)) From the standard - normal table, (P(Z < 1.73)=0.9582) and (P(Z < 0.4)=0.6554) (P(0.4<Z<1.73)=0.9582 - 0.6554=0.3028\approx0.3031)

Answer:

Question #7: A. 28 Question #8: A. 0.3031