Suppose that the scores on a statewide standardized test are normally distributed with a mean of 64 and a standard deviation of 2. Estimate the percentage of scores that were (a) between 60 and 68.

95. 45%

Step-by-step explanation:

Given :

Mean score, = 64

Standard deviation = 2

Score between 60 and 68

P(Z < (x - mean) / standard deviation)

P(Z < (60 - 64) / 2) = P(Z < - 2) = 0.02275 (Z probability calculator)

P(Z < (68 - 64) / 2) = P(Z < 2) = 0.97725 (Z probability calculator)

P( score between 60 and 68)

P(Z < 2) - P(Z < - 2)

0.97725 - 0.02275 = 0.9545

Score between 60 and 68 = 0.9545 = 0.9545 * 100% = 95.45%

part a

in a city ice cream is mostly only bought when the temperature is very high as ice cream cools down our

so we can derive the conclusion that as the temperature increases the number of ice creams sold also increases

part b

y intercept is 5

slope is 2

happy to

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a variable is any characteristics, number, or quantity that can be measured or counted. a variable may also be called a data item. hope i !