B. A soft-drink vending machine is supposed to pour 8 ounces of the drink into a paper cup. However, the actual amount poured into a cup varies. The amount poured into a cup follows a normal distribution with a mean that can be set to any desired amount by adjusting the machine. The standard deviation of the amount poured is always 0.07 ounce regardless of the mean amount. If the owner of the machine wants to be 99% sure that the amount in each cup is 8 ounces or more, to what level should she set the mean?
B. A soft-drink vending machine is supposed to pour 8 ounces of the drink into a paper cup. However, the actual amount poured into a cup varies. The amount poured into a cup follows a normal distribution with a mean that can be set to any desired amount by adjusting the machine. The standard deviation of the amount poured is always 0.07 ounce regardless of the mean amount. If the owner of the machine wants to be 99% sure that the amount in each cup is 8 ounces or more, to what level should she set the mean?
for the 99% ,is it a confinece interval??
the 8 ounce will come under what quantity (n) or (x)??
and is it T-distribution..?
sales manager was interested in determining if there is a relationship between college GPA and sales performance among salespeople hired within the last year. A sample of recently hired salespeople was selected and the number of units each sold last month recorded. The data, scatterplot, regression results, and residual plots appear below.
The regression equation is
Units Sold = - 0.48 + 7.42 GPA
Predictor Coef SE(Coef) T P
Constant -0.484 3.256 -0.15 0.884
GPA 7.423 1.044 7.11 0.000
S = 1.57429 R-Sq = 78.3% R-Sq(adj) = 76.8%
Reject the null
WHAT'S THE NULL???!!!
To determine the level at which the mean should be set, we need to find the cutoff value associated with a 99% confidence level for a normal distribution.
Step 1: Find the z-score for a 99% confidence level.
To find the z-score, we can use a standard normal distribution table or an online calculator. For a 99% confidence level, the z-score is approximately 2.33. This means that 99% of the values in a normal distribution are below 2.33 standard deviations away from the mean.
Step 2: Determine the desired amount.
Since we want to ensure that the amount poured into each cup is 8 ounces or more, the desired amount is 8 ounces.
Step 3: Calculate the mean.
To calculate the mean, we can use the formula: mean = desired amount + (z-score * standard deviation)
mean = 8 + (2.33 * 0.07)
mean = 8 + 0.1631
mean ≈ 8.1631 ounces
Therefore, the owner should set the mean to approximately 8.1631 ounces to be 99% sure that the amount in each cup is 8 ounces or more.